Rs Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 11 T Ratios Of Some Particular Angles are provided here with simple step-by-step explanations. These solutions for T Ratios Of Some Particular Angles are extremely popular among Class 10 students for Maths T Ratios Of Some Particular Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2021 2022 Book of Class 10 Maths Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 554:

Answer:

3x2-1x2=93x2-1x2=139x2-9x2=133x2-3x2
=13cosecθ2-cotθ2=13cosec2θ-cot2θ=131=13

Page No 554:

Answer:

From the given right-angled triangle, we have:
BCAC = sin 30oBC20 = 12 BC = 202 = 10 cm Also, ABAC = cos 30oAB20 = 32 AB = 20×32 = 103  cm BC = 10 cm and AB = 103  cm

Page No 554:

Answer:

From the given right-angled triangle, we have:

BCAB=tan 30o6AB=13  AB=63 cmAlso, BCAC=sin 30o6AC=12  AC=2×6=12 cmAB = 63 cm and AC = 12 cm

Page No 554:

Answer:

From right-angled ∆ABC, we have:

    BCAC=sin 45oBC32=12  BC=3 cm  Also, ABAC=cos 45oAB32=12  AB=3 cm BC=3 cm and AB=3 cm



Page No 572:

Answer:

On substituting the values of various T-ratios, we get:
 sin 60o cos 30o + cos 60o sin 30o 
 =32×32+12×12=34+14=44=1

Page No 572:

Answer:

Ans

Page No 572:

Answer:

On substituting the values of various T-ratios, we get:
 cos 60o cos 30o − sin 60o sin 30o 
 =12×32-32×12=34-34=0

Page No 572:

Answer:

On substituting the values of various T-ratios, we get:
 cos 45o cos 30o + sin 45o  sin 30o 
 = 12×32 + 12×12 = 322 + 122 =3 +122

Page No 572:

Answer:

As we know that,
tan30°=13sec30°=23cosec60°=23tan60°=3By substituting these values, we gettan30° cosec60°+tan60° sec30°=13×23+3×23                                                =23+2                                                =2+63                                                =83Hence, tan30° cosec60°+tan60° sec30°=83
 

Page No 572:

Answer:

On substituting the values of various T-ratios, we get:
 2 cos2 60o + 3 sin2 45o − 3 sin2 30o + 2 cos2 90o
 =2×122+3×122-3×122 + 2×02=2×14+3×12-3×14+0= 12+32-34=2+6-34=54

Page No 572:

Answer:

As we know that,
sin30°=12cos45°=12tan30°=13sin90°=1cot60°=13By substituting these values, we getsin230° cos245°+4tan230°+12sin290°+18cot260°=122×122+4132+1212+18132                                                                          =14×12+413+121+1813                                                                          =18+43+12+124                                                                          =3+32+12+124                                                                          =4824                                                                          =2Hence, sin230° cos245°+4tan230°+12sin290°+18cot260°=2.

Page No 572:

Answer:

On substituting the values of various T-ratios, we get:
  cot2 30o − 2 cos2 30o − 34 sec2 45o + 14 cosec2 30o
 = 32-2×322-34×22+14×22= 3-2×34-34×2+14×4= 3-32-32+1 = 4-32+32 = 4-3= 1

Page No 572:

Answer:

As we know that,
cos0°=1cos45°=12sin30°=12sin90°=1cos60°=12By substituting these values, we getcos0°+cos45°+sin30°sin90°-cos45°+cos60°=1+12+121-12+12                                                                             =1+12+121+12-12                                                                             =1+122-122                                                                             =1+14+2112-12                                                                             =1+14+1-12                                                                             =4+1+4-24                                                                             =74Hence, cos0°+cos45°+sin30°sin90°-cos45°+cos60°=74.

Page No 572:

Answer:

(i)
 LHS=1-sin 60ocos 60o=1-3212=2-3212=2-32×2=2-3RHS= tan 60o-1tan 60o+1=3-13+1=3-13+1×3 -13 -1=3-1232-12=3+1-233-1=4-232=2-3

Hence, LHS = RHS

 1-sin 60ocos 60o=tan 60o-1tan 60o+1

(ii)
 LHS = cos 30o+sin 60o1+sin 30o+cos 60o=32+32 1+12+12 =3+322+1+12=32Also, RHS = cos 30o=32

Hence, LHS = RHS

   cos 30o+sin 60o1+sin 30o+cos 60o=cos 30o1sin60°cos60°

Page No 572:

Answer:

(i) sin 60o cos 30o − cos 60o sin 30o
 =32×32-12×12=34-14=24=12Also, sin 30o=12
∴ sin 60o cos 30o − cos 60o sin 30o = sin 30o

(ii) cos 60o cos 30o + sin 60o sin 30o

=12×32+32×12=34+34=32Also, cos 30o =32
∴ cos 60o cos 30o + sin 60o sin 30o = cos 30o

(iii) 2 sin 30o cos 30o
=2×12×32=32Also, sin 60o =32
∴ 2 sin 30o cos 30o = sin 60o

(iv) 2 sin 45o cos 45o
=2×12×12=1
Also, sin 90o = 1
∴ 2 sin 45o cos 45o = sin 90o

Page No 572:

Answer:

A = 45o
⇒ 2A = 2 × 45o = 90o

(i) sin 2A = sin 90o = 1
2 sin A cos A = 2 sin 45o cos 45o 2×12×12 = 1
∴ sin 2A = 2 sin A cos A

(ii) cos 2A = cos 90o = 0
2 cos2 A − 1 = 2 cos2 45o − 1 =  2×122 - 1 = 2×12 -1 = 1-1 = 0
Now, 1 − 2 sin2 A1-2×122 = 1 - 2×12  =1 - 1 = 0
∴ cos 2A 2 cos2 A − 1 = 1 − 2 sin2 A

Page No 572:

Answer:

A = 30o
⇒ 2A = 2 × 30o = 60o

​(i) sin 2A = sin 60o = 32
2 tan A1+tan2 A=2 tan 30o1+tan2 30o=2×131+132 =231 + 13=2343=23×34=32
sin 2A=2tan A1+tan2A

(ii) cos 2A = cos 60o = 12

1-tan2 A1+tan2 A=1-tan2 30o1+tan2 30o=1-1321+132 =1 - 131 + 13=2343=23×34=12
cos 2A=1-tan2 A1+tan2 A

(iii) tan 2A = tan 60o = 3
2 tan A1-tan2 A=2 tan 30o1-tan2 30o=2×131-132 =231-13=2323=23×32=3
tan 2A=2 tan A1-tan2 A=2tanA1+tan2A

Page No 572:

Answer:

A = 60o and B = 30o
Now, A + B = 60o + 30o​ = 90o
Also, A − B = 60o − 30o = 30o

(i) sin (A + B) = sin 90o = 1
sin A cos B + cos A sin B = sin 60o cos 30o + cos 60o sin 30o
32×32+12×12=34 + 14=1
∴ sin (A + B) = sin A cos B + cos A sin B


(ii) cos (A + B) = cos 90o = 0
cos A cos B − sin A sin B =cos 60o cos 30o − sin 60o sin 30o  =12×32 - 32×12 = 34 - 34 = 0
∴​ cos (A + B) = cos A cos B − sin A sin B
 



Page No 573:

Answer:

(i) sin (A − B) = sin 30o = 12
sin A cos B − cos A sin B = sin 60o cos 30o − cos 60o sin 30o
32× 32- 12×12 = 34 - 14 = 24 = 12
∴ sin (A − B) = sin A cos B − cos A sin B

(ii) cos (AB) = cos 30o = 32
cos A cos B + sin A sin B = cos 60o cos 30o + sin 60o sin 30o
= 12×32 + 32×12 = 34 + 34 = 2×34 = 32
∴​ cos (AB) = cos A cos B + sin A sin B

(iii) tan (AB) = tan 30o = 13
 tan A-tan B1+tan A tan B=tan 60o- tan 30o1 +tan 60o tan 30o= 3 -13 1+3×13=12×3- 13=13
∴​ tan (AB) = tan A-tan B1+tan A tan B

Page No 573:

Answer:

Given:
tan A = 13 and tan B= 12tan (A+B) = tan A+tan B1-tan A tan BOn substituting these values in RHS of the expression, we get:tan A+tan B1-tan A tan B=13+121-13×12=561-16=5656=1 tan (A+B)=1=tan 45o       [ tan 45o =1]A+B=45o

tan A = 13 and tan B = 12By substituting the values, we get;tan (A+ B) = tan A + tan B1- tan A tan B = 13 +121-13×12 = 561-16 = 5656 = 1tan (A + B) = 1 = tan 45oHence, A + B = 4513

Page No 573:

Answer:

A = 30o
⇒ 2A = 2 × 30o = 60o

By substituting the value of the given T-ratio, we get:
tan 2A = 2 tan A1-tan2 A tan 60o=2 tan 30o1-tan2 30o=2×131-132=231-13=2323=23×32=3
∴ tan 60o = 3

Page No 573:

Answer:

A = 30o
⇒ 2A = 2 × 30o = 60o

By substituting the value of the given T-ratio, we get:
cos A=1+cos 2A2 cos 30o=1+ cos 60o2=1+122=322  =34=32
∴ cos 30o = 32

Page No 573:

Answer:

A = 30o
⇒ 2A = 2 × 30o = 60o

By substituting the value of the given T-ratio, we get:

sin A=1-cos 2A2sin 30o=1-cos 60o2=1-122=122  =14=12
∴ sin 30o = 12

Page No 573:

Answer:

Let A=45° and B=30°

iAs, sinA+B=sinA cosB+cosA sinBsin45°+30°=sin45° cos30°+cos45° sin30°sin75°=12×32+12×12sin75°=322+122 sin75°=3+122

iiAs, cosA-B=cosA cosB+sinA sinBcos45°-30°=cos45° cos30°+sin45° sin30°cos15°=12×32+12×12cos15°=322+122 cos15°=3+122

Disclaimer: cos15° can also be calculated by taking A=60° and B=45°.

Page No 573:

Answer:

As we know that,
tan45°=1Thus,if tanx+30°=1x+30°=45°x=45°-30°x=15°Hence, the value of x is 15°.

Page No 573:

Answer:

Given: cotθ-1cotθ+1=1-31+3cotθ-1cotθ+1=1-31+31tanθ-11tanθ+1=1-31+3              cotθ=1tanθ1-tanθtanθ1+tanθtanθ=1-31+31-tanθ1+tanθ=1-31+3On comparing LHS and RHS, we gettanθ=3θ=60°            tan60°=3Hence, the acute angle θ is equal to 60°.

Page No 573:

Answer:

As we know that,
sin90°=1Thus,if sinA+B=1A+B=90°          ...1and tan30°=13Thus,if tanA-B=13A-B=30°          ...2Solving 1 and 2, we getA=60° and B=30°Hence, the values of A and B are 60° and 30°,respectively.

Page No 573:

Answer:

As we know that,
sin60°=32Thus,if sinA+B=32A+B=60°          ...1and cos30°=32Thus,if cosA-B=32A-B=30°          ...2Solving 1 and 2, we getA=45° and B=15°Hence, the values of A and B are 45° and 15°,respectively.

Page No 573:

Answer:

Here, tan (A − B) = 13 
⇒ tan (A B) = tan 30o       [∵ tan 30o = 13]
A − B = 30o                     ...(i)

Also, tan (A + B) = 3 
⇒​ tan (A + B) =  tan 60o        [∵ tan 60o = 3]
A + B = 60o                           ...(ii)  

Solving (i) and (ii), we get:
A = 45o and B = 15o



Page No 574:

Answer:

As we know that,
cosec90°=1Thus,if cosecA+B=1A+B=90°          ...1and cosec30°=2Thus,if cosecA-B=2A-B=30°          ...2Solving 1 and 2, we getA=60° and B=30°           ...3Now,(i) sin60°=32cos30°=32cos60°=12sin30°=12On substituting these values, we getsinA cosB+cosA sinB=sin60° cos30°+cos60° sin30°                                 =32×32+12×12                                 =34+14                                 =44                                 =1Hence, sinA cosB+cosA sinB=1.(ii) tan60°=3tan30°=13On substituting these values, we gettanA-tanB1+tanA tanB=tan60°-tan30°1+tan60° tan30°                      =3-131+313                      =3-131+1                      =223                      =13Hence, tanA-tanB1+tanA tanB=13.

Page No 574:

Answer:

(i) As we know that,
tan45°=1cot60°=13sin30°=12cosec60°=23On substituting these values, we getxtan45° cot60°=sin30° cosec60°x113=1223x3=13x=1Hence, the value of x is 1.


(ii) As we know that,
tan30°=13sin60°=32cosec30°=2On substituting these values, we get2cosec230°+xsin260°-34tan230°=10222+x322-34132=1024+x34-3413=108+3x4-14=103x4-14=10-83x-14=23x-1=83x=9x=3Hence, the value of x is 3.



Page No 576:

Answer:

As we know that,
sec60°=2By substituting the value, we getsec260°-1=22-1                   =4-1                   =3Hence, the correct option is c.

Page No 576:

Answer:

As we know that,
sin30°=12sec60°=2cot45°=1By substituting these values, we getsin230°-sec260°+4cot245°=122-22+412                                            =14-4+4                                            =14Hence, the correct option is d.

Page No 576:

Answer:

As we know that,
sin45°=12cos60°=12cot30°=3By substituting these values, we get3cos260°+2cot230°-5sin245°=3122+232-5122                                                =314+23-512                                                =34+6-52                                                =3+24-104                                                =174Hence, the correct option is c.

Page No 576:

Answer:

As we know that,
cos30°=32cos45°=12sec60°=2cos90°=0tan60°=3By substituting these values, we getcos230° cos245°+4sec260°+12cos290°-2tan260°=322122+422+1202-232                                                                              =3412+44-23                                                                              =38+16-6                                                                              =38+10                                                                              =3+808                                                                              =838Hence, the correct option is b.

Page No 576:

Answer:

As we know that,
cos0°=1cos45°=12sin30°=12sin90°=1cos60°=12sin45°=12By substituting these values, we getcos0°+sin30°+sin45°sin90°+cos60°-cos45°=1+12+121+12-12                                                                            =1+12+121+12-12                                                                            =1+122-122                                                                            =1+14+2112-12                                                                            =1+14+1-12                                                                            =4+1+4-24                                                                            =74Hence, the correct option is a.

Page No 576:

Answer:

As we know that,
cos45°=12cos30°=32tan45°=1sin45°=12By substituting these values, we gettan245°-cos230°=x sin45° cos45°12-322=x12121-34=x24-34=x2x2=14x=12Hence, the correct option is c.

Page No 576:

Answer:

As we know that,
sin45°=12Thus,if 2 sin60°-α=1sin60°-α=1260°-α=45°α=60°-45°α=15°Hence, the correct option is a.

Page No 576:

Answer:

Given: tan= 3cotx


tanx=3cotxtanx=31tanx          cotx=1tanxtan2x=3tanx=3x=60°                   tan60°=3  Hence, the correct option is a.

Page No 576:

Answer:

As we know that,
tan60°=3Thus,if 3 tan2θ-3=03 tan2θ=3tan2θ=33tan2θ=32θ=60°θ=30°Hence, the correct option is b.

Page No 576:

Answer:

As we know that,
sin60°=32Thus,if 2sin2θ=3sin2θ=322θ=60°θ=30°Hence, the correct option is a.

Page No 576:

Answer:

As we know that,
cos60°=32Thus,if 2cos3θ=1cos3θ=123θ=60°θ=20°Hence, the correct option is c.

Page No 576:

Answer:

As we know that,
tan45°=1cos60°=12sin60°=32cot60°=13By substituting these values, we getx tan45° cos60°=sin60° cot60°x112=3213x2=12x=1Hence, the correct option is a.



Page No 577:

Answer:

As we know that,
tan45°=1Thus,if tan3x+30°=13x+30°=45°3x=45°-30°3x=15°x=5°Hence, the correct option is d.

Page No 577:

Answer:

Given: sinθ=cosθsinθ=sin90°-θ                 sin90°-θ=cosθθ=90°-θ2θ=90°θ=45°Hence, the correct option is b.

Page No 577:

Answer:

As we know that,
sin30°=12Thus,if sinA-B=12A-B=30°          ...1and cos60°=12Thus,if cosA+B=12A+B=60°          ...2Solving 1 and 2, we getA=45° and B=15°Hence, the correct optiom is c.



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