Rd Sharma 2020 Solutions for Class 10 Math Chapter 10 Trigonometric Ratios are provided here with simple step-by-step explanations. These solutions for Trigonometric Ratios are extremely popular among Class 10 students for Math Trigonometric Ratios Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2020 Book of Class 10 Math Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2020 Solutions. All Rd Sharma 2020 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.
Page No 10.23:
Question 1:
In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
(i)
(ii)
(iii) tan θ = 11
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
(xii)
Answer:
(i) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Perpendicular side = 2 and
Hypotenuse = 3
Therefore, by Pythagoras theorem,
Now we substitute the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)
Therefore,
Hence, Base =
Now,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(ii) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Base = 4 and
Hypotenuse = 5
Therefore,
By Pythagoras theorem,
Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)
Hence, Perpendicular side = 3
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(iii) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Base = 1 and
Perpendicular side = 5
Therefore,
By Pythagoras theorem,
Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC)
Hence, Hypotenuse =
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(iv) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Perpendicular side = 11 and
Hypotenuse = 15
Therefore,
By Pythagoras theorem,
Now we substitute the value of perpendicular side (BC) and hypotenuse(AC) and get the base side (AB)
Hence, Base =
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(v) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Base = 12 and
Perpendicular side = 5
Therefore,
By Pythagoras theorem,
Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC)
Hence, Hypotenuse = 13
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(vi) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Perpendicular side = and
Hypotenuse = 2
Therefore,
By Pythagoras theorem,
Now we substitute the value of perpendicular side (BC) and hypotenuse(AC) and get the base side (AB)
Hence, Base = 1
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(vii) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Base = 7 and
Hypotenuse = 25
Therefore,
By Pythagoras theorem,
Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)
Hence, Perpendicular side = 24
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(viii) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Base = 15 and
Perpendicular side = 8
Therefore,
By Pythagoras theorem,
Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC)
Hence, Hypotenuse = 17
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(ix) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Base = 12 and
Perpendicular side = 5
Therefore,
By Pythagoras theorem,
Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC)
Hence, Hypotenuse = 13
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(x) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Base = 5 and
Hypotenuse = 13
Therefore,
By Pythagoras theorem,
Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)
Hence, Perpendicular side = 12
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(xi) Given:
…… (1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Perpendicular side = 1 and
Hypotenuse =
Therefore,
By Pythagoras theorem,
Now we substitute the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)
Hence, Base side = 3
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(xii) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Base = 12 and
Hypotenuse = 15
Therefore,
By Pythagoras theorem,
Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)
Hence, Perpendicular side = 9
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Page No 10.23:
Question 2:
In a ∆ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine
(i) sin A, cos A
(ii) sin C, cos C
Answer:
(i) The given triangle is below:-
Given: In ,
To Find:
In this problem, Hypotenuse side is unknown
Hence we first find Hypotenuse side by Pythagoras theorem
By Pythagoras theorem,
We get,
By definition,
By definition,
Answer:
(ii) The given triangle is below:
Given: In ΔABC,
To Find:
In this problem, Hypotenuse side is unknown
Hence we first find Hypotenuse side by Pythagoras theorem
By Pythagoras theorem,
We get,
By definition,
By definition,
Answer:
Page No 10.23:
Question 3:
In the given figure, find tan P and cot R. Is tan P = cot R?
Answer:
The given figure is below:
To Find:
In the given right angled ΔPQR, length of side QR is unknown.
Therefore, to find length of side QR we use Pythagoras Theorem
Hence, by applying Pythagoras theorem in ΔPQR,
We get,
Now, we substitute the length of given side PR and PQ in the above equation
By definition, we know that
… (1)
Also, by definition, we know that
… (2)
Comparing equation (1) and (2) , we come to know that R.H.S of both the equation are equal
Therefore, L.H.S of both the equation are also equal
Answer:
Page No 10.24:
Question 4:
If , compute cos A and tan A.
Answer:
Given: ……(1)
To Find:
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Perpendicular side = 9 and
Hypotenuse = 41
Now using the perpendicular side and hypotenuse we can construct as shown below
Length of side AB is unknown in right angled ,
To find length of side AB, we use Pythagoras theorem.
Therefore, by applying Pythagoras theorem in ,
We get,
Hence, length of side AB = 40
Now,
By definition,
Now,
By definition,
Answer: and
Page No 10.24:
Question 5:
Given 15 cot A = 8, find sin A and sec A.
Answer:
Given: 15 = 8
To Find:
Since
By taking 15 on R.H.S
We get,
By definition,
Hence,
Comparing equation (1) and (2)
We get,
= 8
= 15
can be drawn as shown below using above information
Hypotenuse side AC is unknown.
Therefore, we find side AC of by Pythagoras theorem.
So, by applying Pythagoras theorem to
We get,
Therefore, Hypotenuse = 17
Now by definition,
Substituting values of sides from the above figure
By definition,
Hence,
Answer: and
Page No 10.24:
Question 6:
In ∆PQR, right angled at Q, PQ = 4 cm and RQ = 3 cm. Find the values of sin P, sin R, sec P and sec R.
Answer:
Given:
is right angled at vertex Q.
To find:
Given is as shown below
Hypotenuse side PR is unknown.
Therefore, we find side PR of by Pythagoras theorem
By applying Pythagoras theorem to
We get,
Hence, Hypotenuse = 5
Now by definition,
Substituting values of sides from the above figure
Now by definition,
Substituting values of sides from the above figure
By definition,
By definition,
Substituting values of sides from the above figure
Answer: , , and
Page No 10.24:
Question 7:
If , evaluate :
(i)
(ii) cot2 θ
Answer:
(i) Given:
To evaluate:
…… (1)
We know the following formula
By applying the above formula in the numerator of equation (1) ,
We get,
… (Where a = 1 and b = )
…… (2)
Similarly,
By applying formula in the denominator of equation (1) ,
We get,
… (Where a = 1 and b = )
…… (3)
Substituting the value of numerator and denominator of equation (1) ,from equation (2) and(3)
Therefore,
…… (4)
Since,
Therefore,
Putting the value of and in Equation (4)
We get,
We know that,
Since,
Therefore,
Answer:
(ii) Given:
To evaluate:
Squaring on both sides,
We get,
Answer:
Page No 10.24:
Question 8:
If 3 cot A = 4, check whether or not.
Answer:
Given:
To check whether or not
Dividing by 3 on both sides,
We get,
…… (1)
By definition,
Therefore,
…… (2)
Comparing Equation (1) and (2)
We get,
= 4
= 3
Hence, is as shown in figure below
In, Hypotenuse is unknown
Hence, It can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem in
We get
Hence, Hypotenuse = 5
To check whether or not
We get the values of
By definition,
Substituting the value from Equation (1)
We get,
….… (3)
Now by definition,
…… (4)
Now by definition,
…… (5)
Now we first take L.H.S of Equation
Substituting value of from equation (3)
We get,
Taking L.C.M on both numerator and denominator
We get,
…… (6)
Now we take R.H.S of Equation
Substituting value of and from equation (4) and (5)
We get,
…… (7)
Comparing Equation (6) and (7)
We get,
Answer: Yes
Page No 10.24:
Question 9:
If , find the value of .
Answer:
Given:
…… (1)
Now, we know that
Therefore equation (1) becomes as follows
Now, by applying invertendo
We get,
Now, by applying Compenendo-dividendo
We get,
Therefore,
Page No 10.24:
Question 10:
If 3 tan θ = 4, find the value of .
Answer:
Given:
Therefore,
…… (1)
Now, we know that
Therefore equation (1) becomes
…… (2)
Now, by applying Invertendo to equation (2)
We get,
…… (3)
Now, multiplying by 4 on both sides
We get,
Therefore
Now by applying dividendo in above equation
We get,
…… (4)
Now, multiplying by 2 on both sides of equation (3)
We get,
Therefore
Now by applying componendo in above equation
We get,
…… (5)
Now, by dividing equation (4) by equation (5)
We get,
Therefore,
Therefore, on L.H.S cancels and we get,
Therefore,
Hence,
Page No 10.24:
Question 11:
If 3 cot θ = 2, find the value of .
Answer:
Given:
Therefore,
…… (1)
Now, we know that
Therefore equation (1) becomes
…… (2)
Now, by applying Invertendo to equation (2)
We get,
…… (3)
Now, multiplying by on both sides
We get,
Therefore, 3 cancels out on R.H.S and
We get,
Now by applying dividendo in above equation
We get,
…… (4)
Now, multiplying by on both sides of equation (3)
We get,
Therefore, 2 cancels out on R.H.S and
We get,
Now by applying componendo in above equation
We get,
…… (5)
Now, by dividing equation (4) by equation (5)
We get,
Therefore,
Therefore, on L.H.S cancels out and we get,
Now, by taking 2 in the numerator of L.H.S on the R.H.S
We get,
Therefore, 2 cancels out on R.H.S. and
We get,
Hence,
Page No 10.24:
Question 12:
If , prove that .
Answer:
Given:
…… (1)
Now, we know that
Therefore equation (1) becomes
…… (2)
Now, multiplying by on both sides of equation (2)
We get,
Therefore,
…... (3)
Now by applying dividendo in above equation (3)
We get,
…… (4)
Now by applying componendo in equation (3)
We get,
…… (5)
Now, by dividing equation (4) by equation (5)
We get,
Therefore,
Therefore, and cancels on L.H.S and R.H.S respectively and we get,
Hence, it is proved that
Page No 10.24:
Question 13:
If , show that .
Answer:
Given:
To show that
Now, we know that
Therefore,
Therefore,
…… (1)
Now, we know that
…… (2)
Now, by comparing equation (1) and (2)
We get,
= 5
And
Hypotenuse = 13
Therefore from above figure
Base side
Hypotenuse
Side AB is unknown, It can be determined by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
Therefore by substituting the values of known sides
We get,
Therefore,
Therefore,
…… (3)
Now, we know that
Now from figure (a)
We get,
Therefore,
…… (4)
Now L.H.S. of the equation to be proved is as follows
Substituting the value of and from equation (1) and (4) respectively
We get,
Therefore,
Hence proved that,
Page No 10.24:
Question 14:
If , show that sin θ (1 − tan θ).
Answer:
Given: ……(1)
To show that
Now, we know that …… (2)
Therefore, by comparing equation (1) and (2)
We get,
= 12
And
Hypotenuse = 13
Therefore from above figure
Base side
Hypotenuse
Side AB is unknown and it can be determined by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
Therefore by substituting the values of known sides
We get,
Therefore,
Therefore,
…… (3)
Now, we know that
Now from figure (a)
We get,
Therefore,
…… (4)
Now, we know that
Now from figure (a)
We get,
Therefore,
…… (5)
Now L.H.S. of the equation to be proved is as follows
…… (6)
Substituting the value of and from equation (4) and (5) respectively
We get,
Taking L.C.M inside the bracket
We get,
Therefore,
Now, by opening the bracket and simplifying
We get,
…… (7)
From equation (6) and (7) , it can be shown that
Page No 10.24:
Question 15:
If , show that .
Answer:
Given: ……(1)
To show that
Now, we know that
Since
Therefore,
Therefore,
…… (2)
Comparing Equation (1) and (2)
We get,
= 1
Therefore, Triangle representing angle is as shown below
Hypotenuse AC is unknown and it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
Therefore by substituting the values of known sides
We get,
Therefore,
Therefore,
…… (3)
Now, we know that
Now from figure (a)
We get,
Therefore from figure (a) and equation (3) ,
…… (4)
Now, we know that
Now from figure (a)
We get,
Therefore from figure (a) and equation (3) ,
…… (5)
Now, L.H.S of the equation to be proved is as follows
Substituting the value of and from equation (4) and (5)
We get,
Now by taking L.C.M. in numerator as well as denominator
We get,
Therefore,
Therefore,
Therefore,
Hence proved that
Page No 10.24:
Question 16:
If , show that
Answer:
Given: ……(1)
To show that
Now, we know that
Since ……(2)
Therefore,
Comparing Equation (1) and (2)
We get,
Therefore, Triangle representing angle is as shown below
Hypotenuse AC is unknown and it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
Therefore by substituting the values of known sides
We get,
Therefore,
Therefore,
…… (3)
Now, we know that
Now from figure (a)
We get,
Therefore from figure (a) and equation (3) ,
…… (4)
Now, we know that
Therefore, from equation (4)
We get,
Therefore,
…… (5)
Now, we know that
Now from figure (a)
We get,
Therefore from figure (a) and equation (3) ,
…… (6)
Now, we know that
Therefore, from equation (6)
We get,
Therefore,
…… (7)
Now, L.H.S of the equation to be proved is as follows
Substituting the value of and from equation (6) and (7)
We get,
Now by taking L.C.M. in numerator as well as denominator
We get,
Therefore,
Therefore,
Therefore,
Hence proved that
Page No 10.24:
Question 17:
If , find the value of .
Answer:
Given: ……(1)
To find the value of
Now we know that
Therefore,
Therefore from equation (1)
…… (2)
Also, we know that
Therefore,
Substituting the value of from equation (2)
We get,
Therefore
…… (3)
Also, we know that .
Therefore,
Therefore
Therefore,
Therefore,
…… (4)
Also
Therefore, from equation (4)
We get,
…… (5)
Substituting the value of,,and from equation (2) (3) (4) and (5) respectively in the expression below
We get,
Therefore,
Page No 10.24:
Question 18:
If , find the value of
Answer:
Given: ……(1)
To find the value of
Now, we know the following trigonometric identity
Therefore, by substituting the value of from equation (1) ,
We get,
By taking L.C.M. on the R.H.S,
We get,
Therefore
Therefore
…… (2)
Now, we know that
Therefore,
Therefore
…… (3)
Now, we know the following trigonometric identity
Therefore,
Now by substituting the value of from equation (3)
We get,
Therefore, by taking L.C.M on R.H.S
We get,
Now, by taking square root on both sides
We get,
Therefore,
…… (4)
Substituting the value of and from equation (3) and (4) respectively in the expression below
Therefore,
Therefore,
Page No 10.25:
Question 19:
If , find the value of .
Answer:
Given: ……(1)
To find the value of
Now, we know the following trigonometric identity
Therefore, by substituting the value of from equation (1) ,
We get,
Therefore,
Therefore by taking square root on both sides
We get,
Therefore,
…… (2)
Now, we know that
Therefore by substituting the value of and from equation (2) and (1) respectively
We get,
…… (4)
Now, by substituting the value of and from equation (2) and (4) respectively in the expression below
We get,
Therefore,
Therefore,
Page No 10.25:
Question 20:
If , evaluate .
Answer:
Given: ……(1)
To find the value of
Now, we know the following trigonometric identity
Therefore, by substituting the value of from equation (1) ,
We get,
Therefore,
Now by taking L.C.M
We get,
Therefore by taking square root on both sides
We get,
Therefore,
…… (2)
Now, we know that
Therefore by substituting the value of and from equation (1) and (2) respectively
We get,
…… (3)
Also, we know that
Therefore from equation (4) ,
We get,
Therefore,
…… (4)
Now, by substituting the value of, and from equation (2) , (3) and (4) respectively in the expression below
We get,
Therefore,
Page No 10.25:
Question 21:
If , find that sin θ + cos θ.
Answer:
Given:
…… (1)
To find:
Now we know is defined as follows
…… (2)
Now by comparing equation (1) and (2)
We get
= 24
= 7
Therefore triangle representing angle is as shown below
Side AC is unknown and can be found using Pythagoras theorem
Therefore,
Now by substituting the value of known sides from figure
We get,
Now by taking square root on both sides
We get,
Therefore Hypotenuse side AC = 25 …… (3)
Now we know, is defined as follows
Therefore from figure (a) and equation (3)
We get,
…… (4)
Now we know, is defined as follows
Therefore from figure (a) and equation (3)
We get,
…… (5)
Now we need to find the value of expression
Therefore by substituting the value of andfrom equation (4) and (5) respectively, we get,
Hence
Page No 10.25:
Question 22:
If , find sec θ + tan θ in terms of a and b.
Answer:
Given:
…… (1)
To find:
Now we know, is defined as follows
…… (2)
Now by comparing (1) and (2)
We get,
= a
and
Hypotenuse = b
Therefore triangle representing angle is as shown below
Here side BC is unknown
Now we find side BC by applying Pythagoras theorem to right angled
Therefore,
Now by substituting the value of sides AB and AC from figure (a)
We get,
Therefore,
Now by taking square root on both sides
We get,
Therefore,
Base side …… (3)
Now we know, is defined as follows
Therefore from figure (a) and equation (3)
We get,
…… (4)
Now we know,
Therefore,
Therefore,
…… (5)
Now we know,
Now by substituting the values from equation (1) and (3)
We get,
Therefore,
…… (6)
Now we need to find
Now by substituting the value of and from equation (5) and (6) respectively
We get,
…… (7)
Now we have the following formula which says
Therefore by applying above formula in equation (7)
We get,
Now by substituting in above expression
We get,
Now present in the numerator as well as denominator of above expression gets cancels and we get,
Square root is present in the numerator as well as denominator of above expression
Therefore we can place both numerator as well as denominator under a common square root sign
Therefore,
Page No 10.25:
Question 23:
If 8 tan A = 15, find sin A − cos A.
Answer:
Given:
Therefore,
…… (1)
To find:
Now we know is defined as follows
…… (2)
Now by comparing equation (1) and (2)
We get
= 15
= 8
Therefore triangle representing angle A is as shown below
Side AC is unknown and can be found using Pythagoras theorem
Therefore,
Now by substituting the value of known sides from figure (a)
We get,
Now by taking square root on both sides
We get,
Therefore Hypotenuse side AC = 17 …… (3)
Now we know, is defined as follows
Therefore from figure (a) and equation (3)
We get,
…… (4)
Now we know, is defined as follows
Therefore from figure (a) and equation (3)
We get,
…… (5)
Now we need to find the value of expression
Therefore by substituting the value of andfrom equation (4) and (5) respectively, we get,
Hence
Page No 10.25:
Question 24:
If , show that .
Answer:
Given:
…… (1)
To show that:
Now we know is defined as follows
…… (2)
Now by comparing equation (1) and (2)
We get
= 20
= 21
Therefore triangle representing angle is as shown below
Side AC is unknown and can be found using Pythagoras theorem
Therefore,
Now by substituting the value of known sides from figure (a)
We get,
Now by taking square root on both sides
We get,
Therefore Hypotenuse side AC = 29 …… (3)
Now we know, is defined as follows
Therefore from figure (a) and equation (3)
We get,
…… (4)
Now we know, is defined as follows
Therefore from figure (a) and equation (3)
We get,
…… (5)
Now we need to find the value of expression
Therefore by substituting the value of andfrom equation (4) and (5) respectively, we get,
Now by taking L.C.M on R.H.S of above equation
We get
Now as 10 is present in numerator as well as denominator of R.H.S of above equation, it gets cancelled and we get
Hence
Page No 10.25:
Question 25:
If cosec A = 2, find the value of .
Answer:
Given:
…… (1)
To find:
Now we know is defined as below
Therefore,
Now by substituting the value of from equation (1)
We get,
…… (2)
Now by substituting the value of in the following identity of trigonometry
We get,
Now by taking L.C.M we get
Now by taking square root on both sides
We get,
Therefore,
…… (3)
Now is defined as follows
Now by substituting the value of and from equation (2) and (3) respectively we get,
Therefore,
…… (4)
Now by substituting the value of, and from equation (2) , (3) and (4) respectively we get,
Now by taking L.C.M we get
Now 2 gets cancelled and we get
Now by taking L.C.M, we get,
Now by opening the brackets in the numerator
We get,
Therefore,
Now by taking 2 common
We get,
Now as is present in both numerator as well as denominator, it gets cancelled
Therefore,
Page No 10.25:
Question 26:
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Answer:
Given:
…… (1)
To show:
is as shown in figure below
Now since ……from (1)
Therefore
Now observe that denominator of above equality is same that is AB
Hence only when
Therefore …… (2)
We know that when two sides of a triangle are equal, then angle opposite to the sides are also equal.
Therefore from equation (2)
We can say that
Angle opposite to side AC = Angle opposite to side BC
Therefore,
Hence,
Page No 10.25:
Question 27:
In a ∆ABC, right angled at A, if , find the value of sin B cos C + cos B sin C.
Answer:
Given:
To find:
The givenis as shown in figure below
Side BC is unknown and can be found using Pythagoras theorem
Therefore,
Now by substituting the value of known sides from figure (a)
We get,
Now by taking square root on both sides
We get,
Therefore Hypotenuse side BC = 2 …… (1)
Now
Therefore,
Now by substituting the values from equation (1) and figure (a)
We get,
…… (2)
Now
Therefore,
Now by substituting the values from equation (1) and figure (a)
We get,
…… (3)
Now
Therefore,
Now by substituting the values from equation (1) and figure (a)
We get,
…… (4)
Now by definition,
Therefore,
Now by substituting the value of and from equation (4) and given data respectively
We get,
Now gets cancelled as it is present in both numerator and denominator
Therefore,
…… (5)
Now by substituting the value of and from equation (2) , (3) , (4) and (5) respectively in
We get,
Page No 10.25:
Question 28:
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) for some angle θ.
Answer:
(i) In, is acute an angle
Therefore,
Minimum value of is 0° and
Maximum value of is 90°
We know that and
tan90° = ∞
Therefore the statement that;
“The value of is always less than1” is false
(ii)
In and, is acute angle
Therefore,
Minimum value of is 0°and
Maximum value of is
We know that cos0° = 1 and
cos90° = 0
Now,
Therefore minimum value of is …… (1)
Now,
Therefore maximum value of is …… (2)
Now consider the given value
Here,
This value 2.4 lies in between 1 and
Now from equation (1) and (2) , we can say that the value lies in between minimum value of (that is 1) and maximum value of (that is)
Hence, , for some value of angle A is true
(iii) Cosecant of angle A is defined as
Also, is defined as
Therefore,
…… (1)
And
is defined as …… (2)
Therefore from equation (1) and (2) , it is clear that and (that is cosecant of angle A) are two different trigonometric angles
Hence, is the abbreviation used for cosecant of angle A is False
(iv) cot A is a trigonometric ratio which means cotangent of angle A
Hence, is the product of cot and A is False
(v)
The value
In, is acute an angle
Therefore,
Minimum value of is 0° and
Maximum value of is 90°
We know that and
sin90° = 1
Therefore the value of should lie between 0 and 1 and must not exceed 1
Hence the given value for (that is) is not possible
Therefore, , for some angle = False
Page No 10.25:
Question 29:
If , find the value of .
Answer:
Given: ……(1)
To Find: The value of expression
Now, we know that
…… (2)
Now when we compare equation (1) and (2)
We get,
= 12
And
Hypotenuse = 13
Therefore, Triangle representing angle is as shown below
Base side BC is unknown and it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
Therefore by substituting the values of known sides
We get,
Therefore,
Therefore,
…… (3)
Now, we know that
Now from figure (a)
We get,
Therefore from figure (a) and equation (3) ,
…… (4)
Now we know that,
Therefore, substituting the value of and from equation (1) and (4)
We get,
Therefore 13 gets cancelled and we get
…… (5)
Now we substitute the value of , and from equation (1) , (4) and (5) respectively in the expression below
Therefore,
We get,
Therefore by further simplifying we get,
Now 169 gets cancelled and gets reduced to
Therefore
Therefore the value of is
That is
Page No 10.25:
Question 30:
If , find the value of .
Answer:
Given: ……(1)
To Find:
The value of expression
Now, we know that
…… (2)
Now when we compare equation (1) and (2)
We get,
= 512
And
Hypotenuse = 13
Therefore, Triangle representing angle is as shown below
Perpendicular side AB is unknown and it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
Therefore by substituting the values of known sides
We get,
Therefore,
Therefore,
…… (3)
Now, we know that
Now from figure (a)
We get,
Therefore from figure (a) and equation (3) ,
…… (4)
Now we know that,
Therefore, substituting the value of and from equation (1) and (4)
We get,
Therefore 13 gets cancelled and we get
…… (5)
Now we substitute the value of, and from equation (1) , (4) and (5) respectively in the expression below
Therefore,
We get,
Therefore by further simplifying we get,
Now 169 gets cancelled and gets reduced to
Therefore
Therefore the value of is
That is
Page No 10.25:
Question 31:
If , verify that .
Answer:
Given:
…… (1)
To verify:
…… (2)
Now we know that
Therefore
Now, by substituting the value of from equation (1)
We get,
Therefore,
…… (3)
Now, we know the following trigonometric identity
Therefore,
Now by substituting the value of from equation (3)
We get,
Now by taking L.C.M
We get,
Now, by taking square root on both sides
We get,
Therefore,
…… (4)
Now, we know that
Now by substituting the value of and from equation (3) and (4) respectively
We get,
Therefore
…… (5)
Now from the expression of equation (2)
Now by substituting the value of and from equation (3) and (4)
We get,
Therefore,
Now by taking L.C.M of both numerator and denominator
We get,
…… (6)
Now from the expression of equation (2)
Now by substituting the value of from equation (5)
We get,
Now by taking L.C.M
We get,
Now,
Therefore,
Therefore,
…… (7)
Now by comparing equation (6) and (7)
We get,
Page No 10.25:
Question 32:
If , prove that .
Answer:
Given:
…… (1)
To prove:
…… (2)
By definition,
…… (3)
By Comparing (1) and (3)
We get,
Perpendicular side = 3 and
Hypotenuse = 4
Side BC is unknown.
So we find BC by applying Pythagoras theorem to right angled,
Hence,
Now we substitute the value of perpendicular side (AB) and hypotenuse (AC) and get the base side (BC)
Therefore,
Hence, Base side BC = …… (3)
Now,
Therefore from fig. a and equation (3)
Therefore,
…… (4)
Now,
Therefore from fig. a and equation (1) ,
…… (5)
Now,
Therefore from fig. a and equation (4) ,
…… (6)
Now,
Therefore by substituting the values from equation (1) and (4) ,
We get,
Therefore,
…… (7)
Now by substituting the value of , and from equation (5) ,(6) and (7) respectively in the L.H.S of expression (2) ,
We get,
Therefore,
Hence it is proved that
Page No 10.25:
Question 33:
If , verify that .
Answer:
Given:
…… (1)
To verify:
…… (2)
Now we know that
Therefore
Now, by substituting the value of from equation (1)
We get,
Therefore,
…… (3)
Now, we know the following trigonometric identity
Therefore,
Now by substituting the value of from equation (3)
We get,
Now by taking L.C.M
We get,
Now, by taking square root on both sides
We get,
Therefore,
…… (4)
Now, we know that
Now by substituting the value of and from equation (3) and (4) respectively
We get,
Therefore
…… (5)
Now from the expression of equation (2)
Now by substituting the value of and from equation (3) and (4)
We get,
Therefore,
Now by taking L.C.M of both numerator and denominator
We get,
Therefore,
…… (6)
Now from the expression of equation (2)
Now by substituting the value of from equation (5)
We get,
Now by taking L.C.M
We get,
Therefore
Therefore,
…… (7)
Now by comparing equation (6) and (7)
We get,
Page No 10.26:
Question 34:
If , prove that .
Answer:
Given:
…… (1)
To prove:
Now we know is defined as follows
…… (2)
Now by comparing equation (1) and (2)
We get,
= 3
= 4
Therefore triangle representing angle is as shown below
Side AC is unknown and can be found using Pythagoras theorem
Therefore,
Now by substituting the value of known sides from figure
We get,
Now by taking square root on both sides
We get,
Therefore Hypotenuse side AC = 5 …… (3)
Now we know, is defined as follows
Therefore from figure (a) and equation (3)
We get,
…… (4)
Now we know
Therefore by substituting the value of from equation (4)
We get,
Therefore,
…… (5)
Now we know, is defined as follows
Therefore from figure (a) and equation (3)
We get,
…… (6)
Now we know
Therefore by substituting the value of from equation (6)
We get,
Therefore,
…… (7)
Now, in expression, by substituting the value of andfrom equation (6) and (7) respectively, we get,
L.C.M of 3 and 4 is 12
Now by taking L.C.M in above expression
We get,
Now 12 gets cancelled and we get,
Now
Therefore,
Now 5 gets cancelled and we get,
Therefore, it is proved that
Page No 10.26:
Question 35:
If 3 cos θ − 4 sin θ = 2 cos θ + sin θ, find tan θ.
Answer:
Given:
To find:
Now consider the given expression
Now by dividing both sides of the above expression by
We get,
Now by separating the denominator for each terms
We get,
Now in the above expression present in both numerator and denominator gets cancelled
Therefore,
…… (1)
Now we know that,
Therefore by substituting in equation (1)
We get,
Now by taking on L.H.S
We get,
Therefore,
Hence
Page No 10.26:
Question 36:
If ∠A and ∠P are acute angles such that tan A = tan P, then show that ∠A = ∠P.
Answer:
Given:
To show:
Consider two right angled triangles ABC and PQR such that
Therefore we have,
and
Since it is given that
Therefore,
Now by interchanging position of AB and QR by cross multiplication
We get,
Let (say) …… (1)
Now by cross multiplication
and …… (2)
Now by using Pythagoras theorem in triangles ABC and PQR
We have,
and
Therefore
and
Now
Now using equation (2)
We get,
Now by taking common
We get,
Therefore,
Now gets cancelled
Therefore,
…… (3)
From (1) and (3)
Therefore,
Hence,
Page No 10.41:
Question 1:
Evaluate each of the following
sin 45° sin 30° + cos 45° cos 30°
Answer:
We have,
…… (1)
Now
So by substituting above values in equation (1)
We get,
Therefore,
Page No 10.41:
Question 2:
Evaluate each of the following
sin 60 cos 30° + cos 60° sin 30°
Answer:
We have to find the value of the expression
…… (1)
Now,
So by substituting above values in equation (1)
We get,
Therefore,
Page No 10.41:
Question 3:
Evaluate each of the following
cos 60° cos 45° − sin 60° sin 45°
Answer:
We have to find the value of the following expression
…… (1)
Now,,
So by substituting above values in equation (1)
We get,
Therefore,
Page No 10.41:
Question 4:
Evaluate each of the following
sin2 30° + sin2 45° + sin2 60° + sin2 60° + sin2 90°
Answer:
We have to find
…… (1)
Now,
,, ,
So by substituting above values in equation (1)
We get,
Now by taking denominator 4 together and simplifying
We get,
Now by taking LCM
We get,
Therefore,
Page No 10.41:
Question 5:
Evaluate each of the following
cos2 30° + cos2 45° + cos2 60° + cos2 90°
Answer:
We have to find the following expression
…… (1)
Now,
,, ,
So by substituting above values in equation (1)
We get,
Now by taking denominator 4 together and simplifying
We get,
Now by taking LCM
We get,
Therefore,
Page No 10.41:
Question 6:
Evaluate each of the following
tan2 30° + tan2 60° + tan2 45°
Answer:
We have to find the following expression
…… (1)
Now,
,,
So by substituting above values in equation (1)
We get,
Now by taking LCM
We get,
Therefore,
Page No 10.41:
Question 7:
Evaluate each of the following
2 sin2 30° − 3 cos2 45° + tan2 60°
Answer:
We have to find the following expression
…… (1)
Now,
,,
So by substituting above values in equation (1)
We get,
In the above equation the first term gets reduced to
Therefore,
In the above equation the first term gets reduced to
Therefore,
Therefore,
Page No 10.41:
Question 8:
Evaluate each of the following
sin2 30° cos2 45° + 4 tan2 30° +
Answer:
We have,
…… (1)
Now,
,, ,,
So by substituting above values in equation (1)
We get,
LCM of 8, 3, 2 and 24 is 48
Therefore by taking LCM
We get,
In the above equation the first term gets reduced to
Therefore,
Page No 10.42:
Question 9:
Evaluate each of the following
4 (sin4 60° + cos4 30°) − 3 (tan2 60° − tan2 45°) + 5 cos2 45°
Answer:
We have,
…… (1)
Now,
,, ,
So by substituting above values in equation (1)
We get,
Now, gets reduced to
Therefore,
Now, gets reduced to
Therefore,
Now by taking LCM
We get,
Therefore,
Page No 10.42:
Question 10:
Evaluate each of the following
(cosec2 45° sec2 30°) (sin2 30° + 4 cot2 45° − sec2 60°)
Answer:
We have,
…… (1)
Now,
, , , ,
So by substituting above values in equation (1)
We get,
Now, in above equation 4 cancel 8 and 2 remains
Hence,
Therefore,
Page No 10.42:
Question 11:
Evaluate each of the following
cosec3 30° cos 60° tan3 45° sin2 90° sec2 45° cot 30°
Answer:
We have,
…… (1)
Now,
, , , , ,
So by substituting above values in equation (1)
We get,
Now, 2 gets cancelled and we get,
Page No 10.42:
Question 12:
Evaluate each of the following
Answer:
We have,
…… (1)
Now,
,, ,
So by substituting above values in equation (1)
We get,
Now, in the third term 4 gets cancelled by 2 and 2 remains
Therefore,
Now in the second term, 4 gets cancelled by 2 and 2 remains
Therefore,
Now, LCM of denominator in the above expression is 6
Therefore by taking LCM
We get,
Now in the above expression, gets reduced to
Therefore,
Page No 10.42:
Question 13:
Evaluate each of the following
(cos 0° + sin 45° + sin 30°) (sin 90° − cos 45° + cos 60°)
Answer:
We have,
…… (1)
Now,
,,
So by substituting above values in equation (1)
We get,
Now, LCM of both the product terms in the above expression is
Therefore we get,
Now by rearranging terms in the numerator of above expression
We get,
Now, by applying formula in the numerator of the above expression we get,
…… (2)
Now, we know that
Therefore,
Now, by substituting the above value of in equation (2)
We get,
Now gets reduced to
Therefore,
Hence,
Page No 10.42:
Question 14:
Evaluate each of the following
Answer:
We have,
…… (1)
Now,
,,,
So by substituting above values in equation (1)
We get,
Now, present in the denominator of above expression gets cancelled and we get,
Now by taking LCM in the above expression we get,
Therefore,
Page No 10.42:
Question 15:
Evaluate each of the following
Answer:
We have,
…… (1)
Now,
, ,
So by substituting above values in equation (1)
We get,
Now LCM of denominator of above expression is 6
Therefore by taking LCM we get,
Hence,
Page No 10.42:
Question 16:
Evaluate each of the following
4(sin4 30° + cos2 60°) − 3(cos2 45° − sin2 90°) − sin2 60°
Answer:
We have,
…… (1)
Now,
,, ,
So by substituting above values in equation (1)
We get,
LCM of 16 and 4 in the first term of above expression is 16 and
Similarly LCM of 2 and 1 in the second term of above expression is 2
Therefore,
Now in the second term of the above expression
Therefore,
Now, in the above expression 4 cancels 16 and 4 remains in the denominator of first term
Therefore,
Now by taking LCM =4 in the above expression
We get,
Now, in the above expression gets reduced to 2
Therefore,
Page No 10.42:
Question 17:
Evaluate each of the following
Answer:
We have,
…… (1)
Now,
, , , , ,
So by substituting above values in equation (1)
We get,
Now,
3 gets cancel in numerator and we get,
Now, in the numerator get reduced to 2and we get,
Therefore,
Page No 10.42:
Question 18:
Evaluate each of the following
Answer:
We have,
…… (1)
Now,
,, ,, ,
So by substituting above values in equation (1)
We get,
Now by further simplifying
We get,
Since,
Therefore,
Now, one gets cancelled and
We get,
Now, by taking LCM
We get,
Therefore,
Page No 10.42:
Question 19:
Evaluate each of the following
Answer:
We have,
…… (1)
Now,
,,
So by substituting above values in equation (1)
We get,
Now by taking terms with denominator 2 together and solving
We get,
Now gets reduced to -2
Therefore,
Therefore,
Page No 10.42:
Question 20:
Find the value of x in each of the following :
Answer:
We have,
Since,
Therefore,
Therefore,
Page No 10.42:
Question 21:
Find the value of x in each of the following :
Answer:
We have,
Since,
Therefore,
Therefore,
Page No 10.42:
Question 22:
Find the value of x in each of the following :
Answer:
We have,
Now by cross multiplying we get,
…… (1)
Now we know that
…… (2)
Therefore from equation (1) and (2)
We get,
…… (3)
Since,
…… (4)
Therefore, by comparing equation (3) and (4) we get,
Therefore,
Page No 10.42:
Question 23:
Find the value of x in each of the following :
tan x = sin 45° cos 45° + sin 30°
Answer:
We have,
…… (1)
Now we know that
and
Now by substituting above values in equation (1), we get,
Therefore,
…… (2)
Since,
…… (3)
Therefore by comparing equation (2) and (3)
We get,
Therefore,
Page No 10.42:
Question 24:
Find the value of x in each of the following :
Answer:
We have,
…… (1)
Now we know that
and
Now by substituting above values in equation (1), we get,
Therefore,
…… (2)
Since,
…… (3)
Therefore by comparing equation (2) and (3)
We get,
Therefore,
Page No 10.42:
Question 25:
Find the value of x in each of the following :
cos 2x = cos 60° cos 30° + sin 60° sin 30°
Answer:
We have,
…… (1)
Now we know that
and
Now by substituting above values in equation (1), we get,
Therefore,
Now gets reduced to
Therefore,
…… (2)
Since,
…… (3)
Therefore by comparing equation (2) and (3)
We get,
Therefore,
Page No 10.42:
Question 26:
If θ = 30°, verify that
(i)
(ii)
(iii)
(iv) cos 3θ = 4 cos3 θ − 3 cos θ
Answer:
(i) Given:
…… (1)
To verify:
…… (2)
Now consider LHS of the expression to be verified in equation (2)
Therefore,
Now by substituting the value of from equation (1) in the above expression
We get,
Now by substituting the value of from equation (1) in the expression
We get,
…… (4)
Now by comparing equation (3) and (4)
We get,
Hence
(ii) Given:
…… (1)
To verify:
…… (2)
Now consider right hand side
Hence it is verified that,
(iii) Given:
…… (1)
To verify:
…… (2)
Now consider left hand side of the equation (2)
Therefore,
Now consider right hand side of equation (2)
Therefore,
Hence it is verified that,
(iv) Given:
…… (1)
To verify:
…… (2)
Now consider left hand side of the expression in equation (2)
Therefore
Now consider right hand side of the expression to be verified in equation (2)
Therefore,
Hence it is verified that,
Page No 10.42:
Question 27:
If A = B = 60°, verify that
(i) cos (A − B) = cos A cos B + sin A sin B
(ii) sin (A − B) = sin A cos B − cos A sin B
(iii)
Answer:
(i) Given:
…… (1)
To verify:
…… (2)
Now consider left hand side of the expression to be verified in equation (2)
Therefore,
Now consider right hand side of the expression to be verified in equation (2)
Therefore,
Hence it is verified that,
(ii) Given:
…… (1)
To verify:
…… (2)
Now consider LHS of the expression to be verified in equation (2)
Therefore,
Now by substituting the value of A and B from equation (1) in the above expression
We get,
Hence it is verified that,
(iii) Given:
…… (1)
To verify:
…… (2)
Now consider LHS of the expression to be verified in equation (2)
Therefore,
Now consider RHS of the expression to be verified in equation (2)
Therefore,
Now by substituting the value of A and B from equation (1) in the above expression
We get,
Hence it is verified that,
Page No 10.42:
Question 28:
If A = 30° and B = 60°, verify that
(i) sin (A + B) = sin A cos B + cos A sin B
(ii) cos (A + B) = cos A cos B − sin A sin B
Answer:
(i) Given
and …… (1)
To verify:
…… (2)
Now consider LHS of the expression to be verified in equation (2)
Therefore,
Now consider RHS of the expression to be verified in equation (2)
Therefore;
Hence it is verified that,
(ii) Given:
and …… (1)
To verify:
…… (2)
Now consider LHS of the expression to be verified in equation (2)
Therefore,
Now consider RHS of the expression to be verified in equation (2)
Therefore,
Hence it is verified that,
Page No 10.43:
Question 29:
If sin (A + B) = 1 and cos (A − B) = 1, 0° < A + B ≤ 90°, A ≥ B find A and B.
Answer:
Given:
…… (1)
…… (2)
We know that,
…… (3)
…… (4)
Now by comparing equation (1) and (3)
We get,
…… (5)
Now by comparing equation (2) and (4)
We get,
…… (6)
Now to get the values of A and B, let us solve equation (5) and (6) simultaneously
Therefore by adding equation (5) and (6)
We get,
Therefore,
Hence
Now by subtracting equation (6) from equation (5)
We get,
Therefore,
Hence
Therefore the values of A and B are as follows
and
Page No 10.43:
Question 30:
(i) If find A and B.
(ii) If , then find the values of A and B.
Answer:
(i)
Given:
…… (1)
…… (2)
We know that,
…… (3)
…… (4)
Now by comparing equation (1) and (3)
We get,
…… (5)
Now by comparing equation (2) and (4)
We get,
…… (6)
Now to get the values of A and B, let us solve equation (5) and (6) simultaneously
Therefore by adding equation (5) and (6)
We get,
Therefore,
Hence
Now by subtracting equation (5) from equation (6)
We get,
Therefore,
Hence
Therefore the values of A and B are as follows
and
(ii)
Also,
Adding (1) and (2), we get
Putting in (1), we get
Thus, the values of A and B are 37.5º and 7.5º, respectively.
Page No 10.43:
Question 31:
If find A and B.
Answer:
Given:
…… (1)
…… (2)
We know that,
…… (3)
…… (4)
Now by comparing equation (1) and (3)
We get,
…… (5)
Now by comparing equation (2) and (4)
We get,
…… (6)
Now to get the values of A and B, let us solve equation (5) and (6) simultaneously
Therefore by adding equation (5) and (6)
We get,
Therefore,
Hence
Now by subtracting equation (5) from equation (6)
We get,
Therefore,
Hence
Therefore the values of A and B are as follows
and
Page No 10.43:
Question 32:
In a ∆ABC right angled at B, ∠A = ∠C. Find the values of
(i) sin A cos C + cos A sin C
(ii) sin A sin B + cos A cos B
Answer:
(i) We have drawn the following figure related to given information
To find:
…… (1)
Now we have,
,
,
Now by substituting the above values in equation (1)
We get,
Therefore,
…… (2)
Now in right angled
By applying Pythagoras theorem
We get,
Now, by substituting above value of AC2 in equation (2)
We get,
Now both numerator and denominator contains
Therefore it gets cancelled and 1 remains
Hence
(ii) We have drawn the following figure
e
To find:
…… (1)
Now we know that sum of all the angles of any triangle is 180°
Therefore,
Since and
Therefore,
It is given that
Therefore,
…… (2)
Now we have,
,
,
Now by substituting the above values in equation (1)
We get,
Since
Therefore
Page No 10.43:
Question 33:
Find acute angles A and B, if .
Answer:
Given:
…… (1)
…… (2)
We know that,
…… (3)
…… (4)
Now by comparing equation (1) and (3)
We get,
…… (5)
Now by comparing equation (2) and (4)
We get,
…… (6)
Now to get the values of A and B, let us solve equation (5) and (6) simultaneously
Therefore by subtracting equation (5) from (6)
We get,
Therefore,
Hence
Now by multiplying equation (5) by 2
We get,
…… (7)
Now by subtracting equation (6) from (7)
We get,
Therefore,
Hence
Therefore the values of A and B are as follows and
Page No 10.43:
Question 34:
In ∆PQR, right-angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠P and ∠R.
Answer:
We are given the following information in the form of triangle
To find: and
Now, in
…… (1)
Now we know that
…… (2)
Now by comparing equation (1) and (2)
We get,
…… (3)
Now we have
Now we know that
Therefore,
Now by cross multiplying
We get,
Therefore,
cm …… (4)
Now we know that
Now we know,
…… (6)
Now by comparing equation (5) and (6)
We get,
…… (7)
Hence from equation (3) and (7)
and
Page No 10.43:
Question 35:
If sin (A − B) = sin A cos B − cos A sin B and cos (A − B) = cos A cos B + sin A sin B, find the values of sin 15° and cos 15°.
Answer:
Given:
…… (1)
…… (2)
To find:
The values of and
In this problem we need to find and
Hence to get angle we need to choose the value of A and B such that
So If we choose and
Then we get,
Therefore by substituting and in equation (1)
We get,
Therefore,
…… (3)
Now we know that,
, ,
Now by substituting above values in equation (3)
We get,
Therefore,
…… (4)
Now by substituting and in equation (2)
We get,
Therefore,
…… (5)
Now we know that,
, ,
Now by substituting above values in equation (5)
We get,
Therefore,
…… (6)
Therefore from equation (4) and (6)
Page No 10.43:
Question 36:
In a right triangle ABC, right angled at C, if ∠B = 60° and AB = 15 units. Find the remaining angles and sides.
Answer:
We are given the following triangle with related information
It is required to find , and length of sides AC and BC
is right angled at C
Therefore,
Now we know that sum of all the angles of any triangle is
Therefore,
…… (1)
Now by substituting the values of known angles and in equation (1)
We get,
Therefore,
Therefore,
Now,
We know that,
Now we have,
AB=15 units and
Therefore by substituting above values in equation (2)
We get,
Now by cross multiplying we get,
Therefore,
…… (3)
Now,
We know that,
Now we have,
AB=15 units and
Therefore by substituting above values in equation (4)
We get,
Now by cross multiplying we get,
Therefore,
Hence,
Page No 10.43:
Question 37:
If ∆ABC is a right triangle such that ∠C = 90°, ∠A = 45° and BC = 7 units. Find ∠B, AB and AC.
Answer:
We are given the following information in the form of the triangle
It is required to find and length of sides AB and AC
In
Now we know that sum of all the angles of any triangle is
Therefore,
…… (1)
Now by substituting the values of known angles and in equation (1)
We get,
Therefore,
Therefore,
…… (2)
Now,
We know that,
Now we have,
BC = 7 units and
Therefore by substituting above values in equation (3)
We get,
Now by cross multiplying we get,
Therefore,
…… (4)
Now,
We know that,
……(5)
Now we have,
and
Therefore by substituting above values in equation (5)
We get,
Now by cross multiplying we get,
Therefore,
…… (6)
Therefore,
From equation (2), (4) and (6)
, ,
Page No 10.43:
Question 38:
In a rectangle ABCD, AB = 20 cm, ∠BAC = 60°, calculate side BC and diagonals AC and BD.
Answer:
We have drawn the following figure
Since ABCD is a rectangle
Therefore,
Now, consider
We know that sum of all the angles of any triangle is
Therefore,
…… (1)
Now by substituting the values of known angles and in equation (1)
We get,
Now in
We know that,
Now we have,
AB = 20cm and
Therefore by substituting above values in equation (2)
We get,
Now by cross multiplying we get,
Therefore,
…… (3)
Now in
We know that,
Now we have from equation (3),
AC=40cm and
Therefore by substituting above values in equation (4)
We get,
Now by cross multiplying we get,
Therefore,
…… (5)
Since ABCD is a rectangle
Therefore,
…… (6)
And
…… (7)
Now in
We know that,
Now by substituting the values of sides from equation (6) and (7)
We get,
Since
Therefore,
That is in
…… (8)
Now in
We know that,
From equation (7)and (8)
Since
Therefore,
Now by cross multiplying we get,
Therefore,
…… (9)
Hence from equation (3), (5) and (9)
Page No 10.43:
Question 39:
If A and B are acute angles such that , find A + B.
Answer:
Given:
…… (1)
…… (2)
…… (3)
Now by substituting the value of and from equation (1) and (2) in equation (3)
We get,
Therefore,
…… (3)
Now we know that
…… (4)
Now by comparing equation (3) and (4)
We get,
Page No 10.43:
Question 40:
Prove that .
Answer:
Consider the left hand side.
Now, consider the right hand side.
So, LHS = RHS
Hence proved.
Page No 10.52:
Question 1:
Evaluate the following :
(i)
(ii)
(iii)
(iv)
(v)
Answer:
(i) Given that
Since
Therefore
(ii) Given that
Since
Therefore
(iii) Given that
Since
(iv) We are given that
Since
Therefore
(v) Given that
Since
Therefore
Page No 10.53:
Question 2:
Evaluate the following :
(i)
(ii) cos 48° − sin 42°
(iii)
(iv)
(v)
(vi)
(vii) cosec 31° − sec 59°
(viii) (sin 72° + cos 18°) (sin 72° − cos 18°)
(ix) sin 35° sin 55° − cos 35° cos 55°
(x) tan 48° tan 23° tan 42° tan 67°
(xi) sec50° sin 40° + cos40° cosec 50°
Answer:
(i) We have to find:
Since and
So
So value of is
(ii) We have to find:
Since .So
So value of is
(iii) We have to find:
Since and
So value of is
(iv) We have to find:
Since and
So value of is
(v) We have to find:
Since and
So value of is
(vi) We have to find:
Sinceand
So
So value of is
(vii) We have to find:
Since.So
So value of is
(viii) We have to find:
Since.So
So value of is
(ix) We find:
Sinceand
So value of is
(x) We have to find
Since.So
So value of is
(xi) We find to find
Since, and .So
So value of is
Page No 10.53:
Question 3:
Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°
(i) sin 59° + cos 56°
(ii) tan 65° + cot 49°
(iii) sec 76° + cosec 52°
(iv) cos 78° + sec 78°
(v) cosec 54° + sin 72°
(vi) cot 85° + cos 75°
(vii) sin 67° + cos 75°
Answer:
(i) We have and.So
Thus the desired expression is
(ii) We knowand.So
Thus the desired expression is
(iii) We know thatand.So
Thus the desired expression is
(iv) We knowand
Thus the desired expression is
(v) We know and.So
Thus the desired expression is
(vi) We know thatand.So
Thus the desired expression is
(vii) We know thatand.So
Thus the desired expression is
Page No 10.53:
Question 4:
Express cos 75° + cot 75° in terms of angles between 0° and 30°.
Answer:
Given that:
Hence the correct answer is
Page No 10.53:
Question 5:
If sin 3A = cos (A − 26°), where 3A is an acute angles, find the value of A.
Answer:
We are given 3A is an acute angle
We have:
Hence the correct answer is
Page No 10.53:
Question 6:
If A, B, C are the interior angles of a triangle ABC, prove that
(i)
(ii)
Answer:
(i) We have to prove:
Since we know that in triangle
Proved
(ii) We have to prove:
Since we know that in triangle
Proved
Page No 10.53:
Question 7:
Prove that :
(i) tan 20° tan 35° tan 45° tan 55° tan 70° = 1
(ii) sin 48° sec 42° + cos 48° cosec 42° = 2
(iii)
(iv)
Answer:
We are asked to find the value of
(i) Therefore
Proved
(ii) We will simplify the left hand side
Proved
(iii) We have,
So we will calculate left hand side
Proved
(iv) We have
We will simplify the left hand side
Proved
Page No 10.53:
Question 8:
Prove the following :
(i) sin θ sin (90° − θ) − cos θ cos (90° − θ) = 0
(ii)
(iii)
(iv)
(v) sin (50° − θ) − cos (40° − θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° = 1
Answer:
(i) We have to prove:
Left hand side
=Right hand side
Proved
(ii) We have to prove:
Left hand side
= right hand side
Proved
(iii) We have to prove:
Left hand side
= right hand side
Proved
(iv) We have to prove:
Left hand side
= Right hand side
Proved
(v) We have to prove:
Left hand side
Since .So
=Right hand side
Proved
Page No 10.54:
Question 9:
Evaluate :
(i)
(ii)
(iii)
(iv) tan 35° tan 40° tan 45° tan 50° tan 55°
(v) cosec (65° + θ) − sec (25° − θ) − tan (55° − θ) + cot (35° + θ)
(vi) tan 7° tan 23° tan 60° tan 67° tan 83°
(vii)
(viii)
(ix)
(x)
(xi)
Answer:
We have to evaluate the following values-
(i) We will use the values of known angles of different trigonometric ratios.
(ii) We will use the values of known angles of different trigonometric ratios.
(iii) We will use the properties of complementary angles.
(iv) We will use the properties of complementary angles.
(v) We will use the properties of complementary angles.
(vi) We will use the properties of complementary angles.
(vii) We will use the properties of complementary angles.
(viii) We will use the properties of complementary angles.
(ix) We will use the properties of complementary angles.
(x) We will use the properties of complementary angles.
(xi)
Page No 10.54:
Question 10:
If sin θ = cos (θ − 45°), where θ and θ − 45° are acute angles, find the degree measure of θ.
Answer:
Given that: where and are acute angles
We have to find
Therefore
Page No 10.54:
Question 11:
If A, B, C are the interior angles of a ∆ABC, show that :
(i)
(ii)
Answer:
(i) We have to prove:
Since we know that in triangle
Dividing by 2 on both sides, we get
Proved
(ii) We have to prove:
Since we know that in triangle
Dividing by 2 on both sides, we get
Proved
Page No 10.54:
Question 12:
If 2θ + 45° and 30° − θ are acute angles, find the degree measure of θ satisfying sin (2θ + 45°) = cos (30° − θ).
Answer:
Given that: where and are acute angles
We have to find
So we have
Hence the value of is
Page No 10.54:
Question 13:
If θ is a positive acute such that sec θ = cosec 60°, find the value of 2 cos2 θ − 1.
Answer:
We have: where is positive acute angle
Now we have to find
Put
Hence the value of is
Page No 10.54:
Question 14:
If cos 2θ = sin 4θ, where 2θ and 4θ are acute angles, find the value of θ.
Answer:
We have:
Given in question and are acute angles. We have to find
Now we have
Page No 10.54:
Question 15:
If sin 3 θ = cos (θ − 6°), where 3 θ and θ − 6° are acute angles, find the value of θ.
Answer:
We have: where and are acute angles
We have to find
Now we proceed as to find
Therefore
Page No 10.54:
Question 16:
If sec 4A = cosec (A − 20°), where 4A is an acute angles, find the value of A.
Answer:
Given: and is an acute angle
We have to find
Now
Hence the value of is
Page No 10.54:
Question 17:
If sec 2A = cosec (A − 42°), where 2A is an acute angles, find the value of A.
Answer:
Given: and is an acute angle
We have to find
So we proceed as follows to calculate
Hence the value of is
Page No 10.54:
Question 18:
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Answer:
Given that,
tan2A = cot(A− 18°)
cot(90° − 2A) = cot(A − 18°) (∵ tanθ = cot(90° − θ))
90° − 2A = A − 18°
108° = 3A
A = 36°Page No 10.55:
Question 1:
If θ is an acute angle such that
(a)
(b)
(c)
(d)
Answer:
Given: and we need to find the value of the following expression
We know that:
So we find,
Hence the correct option is
Page No 10.55:
Question 2:
If is equal to
(a)
(b)
(c)
(s)
Answer:
Given:
We have to find the value of following expression in terms of a and b
We know that:
Now we find,
Hence the correct option is
Page No 10.55:
Question 3:
If 5 tan θ − 4 = 0, then the value of is
(a)
(b)
(c) 0
(d)
Answer:
Given that:.We have to find the value of the following expression
Since
We know that:
Since and
Now we find
Hence the correct option is
Page No 10.55:
Question 4:
If 16 cot x = 12, then equals
(a)
(b)
(c)
(d) 0
Answer:
We are given .We are asked to find the following
We know that:
Now we have
,
We knowand
Now we find
Hence the correct option is
Page No 10.55:
Question 5:
If 8 tan x = 15, then sin x − cos x is equal to
(a)
(b)
(c)
(d)
Answer:
Given that:
We know that and
We find:
Hence the correct option is
Page No 10.55:
Question 6:
If
(a)
(b)
(c)
(d)
Answer:
Given that:
We are asked to find the value of the following expression
Since
We know that and
We find:
Hence the correct option is
Page No 10.56:
Question 7:
If , then cos2 θ − sin2 θ =
(a)
(b) 1
(c)
(d)
Answer:
Given that:
Since
We know that and
We find:
Hence the correct option is
Page No 10.56:
Question 8:
If θ is an acute angle such that , then the value of is
(a)
(b)
(c)
(d)
Answer:
Given that: andis an acute angle
We have to find the following expression
Since
Since
We know thatand
We find:
Hence the correct option is
Page No 10.56:
Question 9:
If 3 cos θ = 5 sin θ, then the value of is
(a)
(b)
(c)
(d) None of these
Answer:
We have,
So we can manipulate it as,
So now we can get the values of other trigonometric ratios,
So now we will put these values in the equation,
So the answer is (a).
Page No 10.56:
Question 10:
If tan2 45° − cos2 30° = x sin 45° cos 45°, then x =
(a) 2
(b) −2
(c)
(d)
Answer:
We are given:
We have to find x
We know that
Hence the correct option is
Page No 10.56:
Question 11:
The value of cos2 17° − sin2 73° is
(a) 1
(b)
(c) 0
(d) −1
Answer:
We have:
Hence the correct option is
Page No 10.56:
Question 12:
The value of is
(a)
(b)
(c) 1
(d) 2
Answer:
We have to evaluate the value. The formula to be used,
So,
Now using the properties of complementary angles,
So the answer is
Page No 10.56:
Question 13:
If , then x =
(a) 1
(b) −1
(c) 2
(d) 0
Answer:
We have:
Here we have to find the value of
As we know that
So
Hence the correct option is
Page No 10.56:
Question 14:
If A and B are complementary angles, then
(a) sin A = sin B
(b) cos A = cos B
(c) tan A = tan B
(d) sec A = cosec B
Answer:
Given: and are are complementary angles
Since
Hence the correct option is
Page No 10.56:
Question 15:
If x sin (90° − θ) cot (90° − θ) = cos (90° − θ), then x =
(a) 0
(b) 1
(c) −1
(d) 2
Answer:
We have:
Here we have to find the value of
We know that
Hence the correct option is
Page No 10.56:
Question 16:
If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to
(a) 1
(b)
(c)
(d)
Answer:
Given that:
Here we have to find the value of
We know that
Hence the correct option is
Page No 10.56:
Question 17:
If angles A, B, C to a ∆ABC from an increasing AP, then sin B =
(a)
(b)
(c) 1
(d)
Answer:
Let the angles of a triangleberespectively which constitute an A.P.As we know that sum of all the three angles of a triangle is. So,
So,
Therefore,
Hence,
So answer is
Page No 10.56:
Question 18:
If θ is an acute angle such that sec2 θ = 3, then the value of is
(a)
(b)
(c)
(d)
Answer:
Given that:
We need to find the value of the expression
.So
Here we have to find:
Hence the correct option is
Page No 10.57:
Question 19:
The value of tan 1° tan 2° tan 3° ...... tan 89° is
(a) 1
(b) −1
(c) 0
(d) None of these
Answer:
Here we have to find:
Hence the correct option is
Page No 10.57:
Question 20:
The value of cos 1° cos 2° cos 3° ..... cos 180° is
(a) 1
(b) 0
(c) −1
(d) None of these
Answer:
Here we have to find:
Hence the correct option is
Page No 10.57:
Question 21:
The value of tan 10° tan 15° tan 75° tan 80° is
(a) −1
(b) 0
(c) 1
(d) None of these
Answer:
Here we have to find:
Now
Hence the correct option is
Page No 10.57:
Question 22:
The value of is
(a) 1
(b) − 1
(c) 2
(d) −2
Answer:
We have to find:
So
Hence the correct option is
Page No 10.57:
Question 23:
If θ and 2θ − 45° are acute angles such that sin θ = cos (2θ − 45°), then tan θ is equal to
(a) 1
(b) −1
(c)
(d)
Answer:
Given that: and are acute angles
We have to find
Where and are acute angles
Since
Now
Put
Hence the correct option is
Page No 10.57:
Question 24:
If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2 sin 3θ − is equal to
(a) 1
(b) 0
(c) −1
(d)
Answer:
We are given that and are acute angles satisfying the following condition
.We are asked to find
Where and are acute angles
Now we have to find:
Hence the correct option is
Page No 10.57:
Question 25:
If A + B = 90°, then is equal to
(a) cot2 A
(b) cot2 B
(c) −tan2 A
(d) −cot2 A
Answer:
We have:
We have to find the value of the following expression
So
Hence the correct option is
Page No 10.57:
Question 26:
is equal to
(a) sin 60°
(b) cos 60°
(c) tan 60°
(d) sin 30°
Answer:
We have to find the value of the following expression
Hence the correct option is
Page No 10.57:
Question 27:
is equal to
(a) tan 90°
(b) 1
(c) sin 45°
(d) sin 0°
Answer:
We have to find the value of the following
So
We know that
Hence the correct option is
Page No 10.57:
Question 28:
Sin 2A = 2 sin A is true when A =
(a) 0°
(b) 30°
(c) 45°
(d) 60°
Answer:
We are given
So
Hence the correct option is
Page No 10.57:
Question 29:
is equal to
(a) cos 60°
(b) sin 60°
(c) tan 60°
(d) sin 30°
Answer:
We are asked to find the value of the following
We know that
Hence the correct option is
Page No 10.57:
Question 30:
If A, B and C are interior angles of a triangle ABC, then
(a)
(b)
(c)
(d)
Answer:
We know that in triangle
So
Hence the correct option is
Page No 10.57:
Question 31:
If , then 2 sec2 θ + 2 tan2 θ − 7 is equal to
(a) 1
(b) 0
(c) 3
(d) 4
Answer:
Given that:
We have to find
As we are given
We know that:
Now we have to find: .So
Hence the correct option is
Page No 10.58:
Question 32:
tan 5° ✕ tan 30° ✕ 4 tan 85° is equal to
(a)
(b)
(c) 1
(d) 4
Answer:
We have to find
We know that
So
Hence the correct option is
Page No 10.58:
Question 33:
The value of + cot 1° cot 2° cot 3° .... cot 90°, is
(a) −2
(b) 2
(c) 1
(d) 0
Answer:
We have to find the value of the following expression
Hence the correct option is
Page No 10.58:
Question 34:
In Fig. 5.47, the value of cos ϕ is
(a)
(b)
(c)
(d)
Answer:
We should proceed with the fact that sum of angles on one side of a straight line is.
So from the given figure,
So, …… (1)
Now from the triangle,
Now we will use equation (1) in the above,
Therefore,
So the answer is
Page No 10.58:
Question 35:
In Fig. 5.48, AD = 4 cm, BD = 3 cm and CB = 12 cm, find the cot θ.
(a)
(b)
(c)
(d)
Answer:
We have the following given data in the figure,
Now we will use Pythagoras theorem in,
Therefore,
So the answer is
Page No 10.58:
Question 1:
The value of (sin30° + cos30°) – (sin60° + cos60°) is ________.
Answer:
The value of (sin30° + cos30°) – (sin60° + cos60°) is ___0___.
Page No 10.58:
Question 2:
The value of is ______.
Answer:
The value of is ___1___.
Page No 10.59:
Question 3:
The value of (sin 45° + cos 45°)3 is _______.
Answer:
The value of (sin 45° + cos 45°)3 is .
Page No 10.59:
Question 4:
The value of (sin 30° + cos 30°)2 – (sin 60° – cos 60°)2 is __________.
Answer:
The value of (sin 30° + cos 30°)2 – (sin 60° – cos 60°)2 is .
Page No 10.59:
Question 5:
The value of (cos2 23° – sin2 67°) is ____________.
Answer:
We know that,
The value of (cos2 23° – sin2 67°) is ______0______.
Page No 10.59:
Question 6:
The value of the expression is ________.
Answer:
We know that,
The value of the expression is ____2____.
Page No 10.59:
Question 7:
Given that sin α = and cos β = , then α + β = _________.
Answer:
Also,
Given that sin α = and cos β = , then α + β = ____90°____.
Page No 10.59:
Question 8:
Given that sin (α – β ) = and cos (α + β) = , then α = _______ β = _________.
Answer:
Also,
Adding (1) and (2), we get
Putting in (2), we get
Given that sin (α – β ) = and cos (α + β) = , then α = ___3___ β = ____45º___.
Page No 10.59:
Question 9:
If 4 tan θ = 3, then is equal to _________.
Answer:
(Dividing numerator and denominator by cosθ)
If 4 tan θ = 3, then is equal to .
Page No 10.59:
Question 10:
If cos 5θ = sinθ and 5θ < 90°, then the value of tan 3θ is ___________.
Answer:
If cos 5θ = sinθ and 5θ < 90°, then the value of tan 3θ is ____1____.
Page No 10.59:
Question 11:
The value of sec2 60° – tan2 60° is __________.
Answer:
The value of sec2 60° – tan2 60° is _____1_____.
Page No 10.59:
Question 12:
If A + B = 90°, then the value of tan2 A – cot2 B is _________.
Answer:
If A + B = 90°, then the value of tan2 A – cot2 B is ____0____.
Page No 10.59:
Question 13:
The value of cos 1° cos 2° cos 3° ...... cos 120° is _______.
Answer:
The value of cos 1° cos 2° cos 3° ...... cos 120° is ___0___.
Page No 10.59:
Question 14:
If tan θ + cot θ = 2 and, 0° < θ < 90° then tan10 θ + cot10 θ is equal to _________.
Answer:
So,
If tan θ + cot θ = 2 and, 0° < θ < 90° then tan10 θ + cot10 θ is equal to ____2____.
Page No 10.59:
Question 15:
If sin θ + cos θ = 1 and 0° ≤ θ ≤ 90°, then the possible values of θ are ________.
Answer:
The given equation is sinθ + cosθ = 1.
When θ = 0°,
LHS = sinθ + cosθ = sin0° + cos0° = 0 + 1 = 1 = RHS
When θ = 90°,
LHS = sinθ + cosθ = sin90° + cos90° = 1 + 0 = 1 = RHS
Thus, the possible values of θ (0° ≤ θ ≤ 90°) satisfying the given equation are 0° and 90°.
If sin θ + cos θ = 1 and 0° ≤ θ ≤ 90°, then the possible values of θ are _0° and 90°_.
Page No 10.59:
Question 16:
If , then cot 2A = _________.
Answer:
Also,
Adding (1) and (2), we get
If , then cot 2A = ____0____.
Page No 10.59:
Question 17:
If ΔABC is an isosceles right triangle right angled at B, then
Answer:
It is given that ΔABC is an isosceles right triangle right angled at B.
.....(1)
In ΔABC,
(Angle sum property of a triangle)
= 1
If ΔABC is an isosceles right triangle right angled at B, then
Page No 10.59:
Question 18:
If α + β = 90° and , then tan α tan β = ________.
Answer:
Now,
If α + β = 90° and , then tan α tan β = ____1____.
Page No 10.59:
Question 19:
If in a triangle ABC, angles A and B are complementary, then the value of cot C is ________.
Answer:
It is given that angles A and B are complementary.
.....(1)
In ∆ABC,
(Angle sum property of a triangle)
If in a triangle ABC, angles A and B are complementary, then the value of cot C is ____0____.
Page No 10.59:
Question 20:
The value of tan 25° tan 10° tan 80° tan 65° is __________.
Answer:
The value of tan 25° tan 10° tan 80° tan 65° is _____1_____.
Page No 10.59:
Question 1:
Write the maximum and minimum values of sin θ.
Answer:
The maximum value of is and the minimum value of is because value of lies between −1 and 1
Page No 10.59:
Question 2:
Write the maximum and minimum values of cos θ.
Answer:
The maximum value of is and the minimum value of is because value of lies between −1 and 1
Page No 10.60:
Question 3:
What is the maximum value of ?
Answer:
The maximum value of is because the maximum value of is that is
Page No 10.60:
Question 4:
What is the maximum value of ?
Answer:
The maximum value of is because the maximum value of is that is
Page No 10.60:
Question 5:
If , find the value of .
Answer:
It is given that .
We have to find .
Page No 10.60:
Question 6:
If , find the value of .
Answer:
Given in question:
We have to find
Hence the value of is
Page No 10.60:
Question 7:
If 3 cot θ = 4, find the value of .
Answer:
We have:
Since we know that in right angle triangle
Now, we find
Hence the value of is
Page No 10.60:
Question 8:
Given , what is the value of ?
Answer:
Given: ,
We know that:
Now we find,
Hence the value of is
Page No 10.60:
Question 9:
If , write the value of .
Answer:
Given:
Now we find,
Hence the value of is
Page No 10.60:
Question 10:
If , then what is the value of cot B?
Answer:
Given that:
Hence the value of is
Page No 10.60:
Question 11:
If A + B = 90° and , what is the value of sin A?
Answer:
We have:
Hence the value of is
Page No 10.60:
Question 12:
Write the acute angle θ satisfying .
Answer:
We have:
Hence the acute angle is
Page No 10.60:
Question 13:
Write the value of cos 1° cos 2° cos 3° ....... cos 179° cos 180°.
Answer:
Given that:
Hence the value of is
Page No 10.60:
Question 14:
Write the value of tan 10° tan 15° tan 75° tan 80°?
Answer:
We have to find:
Hence the value of is
Page No 10.60:
Question 15:
If A + B = 90° and , what is cot B?
Answer:
Given in question:
Hence the value of is
Page No 10.60:
Question 16:
If , find the value of (sin A + cos A) sec A.
Answer:
Given:
We know that:
Now we find,
Hence the value of is
Page No 10.60:
Question 17:
In the given figure, PS = 3 cm, QS = 4 cm, ∠PRQ = θ, ∠PSQ = 90°, PQ ⊥ RQ and RQ = 9 cm. Evaluate tan θ.
Answer:
In right ∆PQS,
In right ∆PQR,
Thus, the value of tanθ is .
Page No 10.60:
Question 18:
Find A, if tan 2A = cot(A – 24°).
Answer:
Thus, the value of A is 38º.
Page No 10.60:
Question 19:
Find the value of sin233° + sin257°.
Answer:
We have
Thus, the value of sin233° + sin257° is 1.
Page No 10.60:
Question 20:
Evaluate sin2 60° + 2tan 45° – cos2 30°.
Answer:
Page No 10.60:
Question 21:
If sin , calculate sec A.
Answer:
We know
Now,
View NCERT Solutions for all chapters of Class 10