Rd Sharma 2022 for Class 10 Maths Chapter 11 - Trigonometric Identities

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Page No 529:

Question 1:

Prove the following trigonometric identities.

(1 − cos² A) cosec² A = 1

Answer:

We know that,

So,

Page No 529:

Question 2:

Prove the following trigonometric identities.

(1 + cot² A) sin² A = 1

Answer:

We know that,

So,

Page No 529:

Question 3:

Prove the following trigonometric identities.

(sec² θ − 1) (cosec² θ − 1) = 1

Answer:

We know that,

So,

Page No 529:

Question 4:

Prove the following trigonometric identities.

$\tan θ + \frac{1}{\tan θ} = \sec θ cosec θ$

Answer:

We know that,

So,

Page No 529:

Question 5:

Prove the following trigonometric identities.

$\frac{\cos θ}{1 + \sin θ} = \frac{1 - \sin θ}{\cos θ}$

Answer:

We know that,

Multiplying the both numerator and the denominator by, we have

Page No 529:

Question 6:

Prove the following trigonometric identities.

$\cos^{2} A + \frac{1}{1 + \cot^{2} A} = 1$

Answer:

We know that,

So,

Page No 530:

Question 7:

Prove the following trigonometric identities.

$\frac{1 - \cos θ}{\sin θ} = \frac{\sin θ}{1 + \cos θ}$

Answer:

We have to prove .

We know that,

Multiplying both numerator and denominator by , we have

Page No 530:

Question 8:

Prove the following trigonometric identities.

$\frac{\sin θ}{1 - \cos θ} = cosec θ + \cot θ$

Answer:

We have to prove .

We know that,

Multiplying both numerator and denominator by , we have

Page No 530:

Question 9:

Prove the following trigonometric identities.

$\frac{1 - \sin θ}{1 + \sin θ} = {(\sec θ - \tan θ)}^{2}$

Answer:

We have to prove

We know that, .

Multiplying both numerator and denominator by , we have

Page No 530:

Question 10:

Prove the following trigonometric identities.

$\frac{(1 + \cot^{2} θ) \tan θ}{\sec^{2} θ} = \cot θ$

Answer:

We have to prove

We know that,

So,

Page No 530:

Question 11:

Prove the following trigonometric identities.

tan²θ − sin²θ = tan²θ sin²θ

Answer:

We have to prove

We know that,

So,

Page No 530:

Question 12:

Prove the following trigonometric identities.

(1 + tan²θ) (1 − sinθ) (1 + sinθ) = 1

Answer:

We have to prove

We know that,

So,

Page No 530:

Question 13:

Prove the following trigonometric identities.

sin² A cot² A + cos² A tan² A = 1

Answer:

We have to prove

We know that,

So,

Page No 530:

Question 14:

Prove the following trigonometric identities.

$\frac{1 + \sin θ}{\cos θ} + \frac{\cos θ}{1 + \sin θ} = 2 \sec θ$

Answer:

We have to prove

We know that,

Multiplying the denominator and numerator of the second term by , we have

Page No 530:

Question 15:

Prove the following trigonometric identities.

$\frac{{(1 + \sin θ)}^{2} + {(1 - \sin θ)}^{2}}{2 \cos^{2} θ} = \frac{1 + \sin^{2} θ}{1 - \sin^{2} θ}$

Answer:

We have to prove that .

We know that,

So,

Page No 530:

Question 16:

Prove the following trigonometric identities.

$\frac{1 + \tan^{2} θ}{1 + \cot^{2} θ} = {(\frac{1 - \tan θ}{1 - \cot θ})}^{2} = \tan^{2} θ$

Answer:

We have to prove

Consider the expression

Again, we have

Page No 530:

Question 17:

Prove the following trigonometric identities.

$\frac{1 + \sec θ}{\sec θ} = \frac{\sin^{2} θ}{1 - \cos θ}$

Answer:

We have to prove

We know that,

Multiplying the numerator and denominator by , we have

Page No 530:

Question 18:

Prove the following trigonometric identities.

sec⁶θ = tan⁶θ + 3 tan²θ sec²θ + 1

Answer:

We need to prove

Solving the L.H.S, we get

Further using the identity, we get

Hence proved.

Page No 530:

Question 19:

Prove the following trigonometric identities.

$\frac{\sec A - \tan A}{\sec A + \tan A} = \frac{\cos^{2} A}{{(1 + \sin A)}^{2}}$

Answer:

We need to prove

Here, we will first solve the LHS.

Now, using , we get

Further, multiplying both numerator and denominator by , we get

Now, using the property, we get

So,

= RHS

Hence proved.

Page No 530:

Question 20:

Prove the following trigonometric identities.

(cosecθ + sinθ) (cosecθ − sinθ) = cot²θ + cos²θ

Answer:

We have to prove

We know that,

So,

Page No 530:

Question 21:

Prove the following trigonometric identities.

(i) $\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$

(ii)

Answer:

(i) We need to prove

Here, rationalising the L.H.S, we get

Further using the property, , we get

So,

Hence proved.

(ii) We need to prove

Here, rationaliaing the L.H.S, we get

Further using the property, , we get

So,

Hence proved.

Page No 530:

Question 22:

Prove that:

(i) $\sqrt{\frac{\sec θ - 1}{\sec θ + 1}} + \sqrt{\frac{\sec θ + 1}{\sec θ - 1}} = 2 cosec θ$

(ii) $\sqrt{\frac{1 + \sin θ}{1 - \sin θ}} + \sqrt{\frac{1 - \sin θ}{1 + \sin θ}} = 2 \sec θ$

Answer:

(i) We have,

(ii) We have,

Page No 530:

Question 23:

Prove the following trigonometric identities.

$\frac{\tan^{2} A}{1 + \tan^{2} A} + \frac{\cot^{2} A}{1 + \cot^{2} A} = 1$

Answer:

In the given question, we need to prove .

Here, we will first solve the LHS.

Now, using , we get

On further solving by taking the reciprocal of the denominator, we get,

Hence proved.

Page No 530:

Question 24:

Prove the following trigonometric identities.

$\frac{1 + \cos θ - \sin^{2} θ}{\sin θ (1 + \cos θ)} = \cot θ$

Answer:

In the given question, we need to prove .

Using the property , we get

So,

$\frac{1 + \cos θ - \sin^{2} θ}{\sin θ (1 + \cos θ)} = \frac{1 + \cos θ - (1 - \cos^{2} θ)}{\sin θ (1 + \cos θ)} = \frac{\cos θ + \cos^{2} θ}{\sin θ (1 + \cos θ)}$

Solving further, we get

Hence proved.

Page No 530:

Question 25:

Prove the following trigonometric identities.

tan²A + cot² A = sec² A cosec² A − 2

Answer:

In the given question, we need to prove

Now, using and in L.H.S, we get

Further, using the identity, we get

Since L.H.S = R.H.S

Hence proved.

Page No 530:

Question 26:

$\frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A} = 2 cosec A$

Answer:

Consider the LHS.
$\frac{\tan A}{1 + \sec A} - \frac{\tan A}{1 - \sec A} = \frac{\tan A (1 - \sec A) - \tan A (1 + \sec A)}{(1 + \sec A) (1 - \sec A)} = \frac{\tan A - \tan A \sec A - \tan A - \tan A \sec A}{(1 - \sec^{2} A)} = \frac{- 2 \tan A \sec A}{(1 - \sec^{2} A)} = \frac{- 2 \tan A \sec A}{- \tan^{2} A} = \frac{2 \sec A}{\tan A} = 2 cosec A$
= RHS
Hence proved.

Page No 530:

Question 27:

Prove the following trigonometric identities.

$1 + \frac{\cot^{2} θ}{1 + cosec θ} = cosec θ$

Answer:

In the given question, we need to prove

Using and, we get

Further, using the property , we get

Hence proved.

Page No 530:

Question 28:

Prove the following trigonometric identities.

$\frac{\cos θ}{cosec θ + 1} + \frac{\cos θ}{cosec θ - 1} = 2 \tan θ$

Answer:

In the given question, we need to prove

Using the identity, we get

Further, using the property , we get

Hence proved.

Page No 530:

Question 29:

Prove the following trigonometric identities.

(cosecA − sinA) (secA − cosA) (tanA + cotA) = 1

Answer:

We have to prove

We know that,

So,

Page No 530:

Question 30:

Prove the following trigonometric identities.

$(1 + \tan^{2} A) + (1 + \frac{1}{\tan^{2} A}) = \frac{1}{\sin^{2} A - \sin^{4} A}$

Answer:

We need to prove .

Using the property , we get

Now, using , we get

Further, using the property, , we get

Hence proved.

Page No 530:

Question 31:

Prove the following trigonometric identities.

cot² A cosec² B − cot² B cosec² A = cot² A − cot² B

Answer:

We have to prove

We know that,

So,

Hence proved.

Page No 530:

Question 32:

Prove the following trigonometric identities.

If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x² − y² = a² − b²

Answer:

Given that,

We have to prove

We know that,

So,

Hence proved.

Page No 530:

Question 33:

Prove the following trigonometric identities.

If cosec θ + cot θ = m and cosec θ − cot θ = n, prove that mn = 1

Answer:

Given:

We have to prove

We know that,

Multiplying the two equations, we have

Hence proved.

Page No 530:

Question 34:

Prove the following trigonometric identities.

(i) $\frac{1 + \cos θ + \sin θ}{1 + \cos θ - \sin θ} = \frac{1 + \sin θ}{\cos θ}$
(ii) $\frac{\sin θ - \cos θ + 1}{\sin θ + \cos θ - 1} = \frac{1}{\sec θ - \tan θ}$
(iii) $\frac{\cos θ - \sin θ + 1}{\cos θ + \sin θ - 1} = cosec θ + \cot θ$
(iv) $(\sin θ + \cos θ) (\tan θ + \cot θ) = \sec θ + cosec θ$

Answer:

(i) We have to prove the following identity-

Consider the LHS.

$\frac{1 + \cos θ + \sin θ}{1 + \cos θ - \sin θ}$

= RHS

Hence proved.

(ii) We have to prove the following identity-

Consider the LHS.

$\frac{\sin θ - \cos θ + 1}{\sin θ + \cos θ - 1}$

(Divide numerator and denominator by)

RHS

Hence proved.

(iii) We have to prove the following identity-

$\frac{\cos θ - \sin θ + 1}{\cos θ + \sin θ - 1} = cosec θ + \cot θ$

Consider the LHS.

$\frac{\cos θ - \sin θ + 1}{\cos θ + \sin θ - 1}$

$= \frac{\cos θ - \sin θ + 1}{\cos θ + \sin θ - 1} \times \frac{\cos θ + \sin θ + 1}{\cos θ + \sin θ + 1} = \frac{{(\cos θ + 1)}^{2} - {(\sin θ)}^{2}}{{(\cos θ + \sin θ)}^{2} - {(1)}^{2}} = \frac{\cos^{2} θ + 1 + 2 \cos θ - \sin^{2} θ}{\cos^{2} θ + \sin^{2} θ + 2 \cos θ \sin θ - 1} = \frac{\cos^{2} θ + 1 + 2 \cos θ - (1 - \cos^{2} θ)}{1 + 2 \cos θ \sin θ - 1} = \frac{2 \cos^{2} θ + 2 \cos θ}{2 \cos θ \sin θ} = \frac{2 \cos θ (\cos θ + 1)}{2 \cos θ \sin θ} = \frac{\cos θ + 1}{\sin θ} = \frac{\cos θ}{\sin θ} + \frac{1}{\sin θ} = \cot θ + cosec θ$

= RHS

Hence proved.

(iv)
Consider the LHS.
$(\sin θ + \cos θ) (\tan θ + \cot θ) = (\sin θ + \cos θ) (\frac{\sin θ}{\cos θ} + \frac{\cos θ}{\sin θ}) = (\sin θ + \cos θ) (\frac{\sin^{2} θ + \cos^{2} θ}{\sin θ \times \cos θ}) = \frac{\sin θ + \cos θ}{\sin θ \times \cos θ} [\sin^{2} θ + \cos^{2} θ = 1] = \frac{1}{\cos θ} + \frac{1}{\sin θ} = \sec θ + cosec θ$
= RHS
Hence proved.

Page No 530:

Question 35:

Prove the given identities:

$\frac{1 + \sec θ - \tan θ}{1 + \sec θ + \tan θ} = \frac{1 - \sin θ}{\cos θ}$

Answer:

(ii)Consider the LHS.

$\frac{1 + \sec θ - \tan θ}{1 + \sec θ + \tan θ} = \frac{(\sec θ - \tan θ) + 1}{1 + \sec θ + \tan θ} = \frac{(\sec θ - \tan θ) + (\sec^{2} θ - \tan^{2} θ)}{1 + \sec θ + \tan θ} (\sec^{2} θ - \tan^{2} θ = 1) = \frac{(\sec θ - \tan θ) (1 + \sec θ + \tan θ)}{1 + \sec θ + \tan θ} = \sec θ - \tan θ = \frac{1}{\cos θ} - \frac{\sin θ}{\cos θ} = \frac{1 - \sin θ}{\cos θ}$
= RHS

Hence proved.

Page No 531:

Question 36:

Prove the following trigonometric identities.

$(1 + \cot A + \tan A) (\sin A - \cos A) = \frac{\sec A}{{cosec}^{2} A} - \frac{cosec A}{\sec^{2} A} = \sin A \tan A - \cot A \cos A$

Answer:

We have prove that

We know that,

So,

Now,

Hence proved.

Page No 531:

Question 37:

Prove the following trigonometric identities.

$\frac{\tan^{3} θ}{1 + \tan^{2} θ} + \frac{\cot^{3} θ}{1 + \cot^{2} θ} = \sec θ cosec θ - 2 \sin θ \cos θ$

Answer:

In the given question, we need to prove

Using the property and, we get

Taking the reciprocal of the denominator, we get

Further, using the identity, we get

Hence proved.

Page No 531:

Question 38:

Prove the following trigonometric identities.

$\frac{\tan A}{{(1 + \tan^{2} A)}^{2}} + \frac{\cot A}{(1 + \cot^{2} A)} = \sin A \cos A$

Answer:

We have to prove

We know that, .

So,

Hence proved.

Page No 531:

Question 39:

Prove the given identities:

${(\frac{1 + \sin θ - \cos θ}{1 + \sin θ + \cos θ})}^{2} = \frac{1 - \cos θ}{1 + \cos θ}$

Answer:

(i) In the given question, we need to prove

Taking common from the numerator and the denominator of the L.H.S, we get

Now, using the property , we get

Using , we get

Taking common from the numerator, we get

Using and , we get

Now, using the property , we get

Hence proved.

Page No 531:

Question 40:

Prove the following trigonometric identities.

If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ − 3 cos θ = ± 3.

Answer:

Given:

We have to prove that 5sin θ – 3cos θ = ±3.

We know that,

Squaring the given equation, we have

⇒ (5sin θ – 3cos θ)² = 9
⇒ 5sin θ – 3cos θ = ±3

Hence proved.

Page No 531:

Question 41:

If $\sin θ + 2 \cos θ = 1$ prove that $2 \sin θ - \cos θ = 2$ .

Answer:

It is given that,
$\sin θ + 2 \cos θ = 1 \Rightarrow 2 \cos θ = 1 - \sin θ . . . . . (i) Squaring both sides, we get \Rightarrow {(2 \cos θ)}^{2} = {(1 - \sin θ)}^{2} \Rightarrow 4 \cos^{2} θ = 1 + \sin^{2} θ - 2 \sin θ \Rightarrow 4 \cos^{2} θ = 1 + \sin^{2} θ - 2 \sin θ - 1 + 1 \Rightarrow 4 \cos^{2} θ = 2 - 2 \sin θ - (1 - \sin^{2} θ) \Rightarrow 4 \cos^{2} θ = 2 - 2 \sin θ - \cos^{2} θ \Rightarrow 5 \cos^{2} θ = 2 (1 - sinθ) \Rightarrow 5 \cos^{2} θ = 4 \cos θ [Using (i)] \Rightarrow \cos θ (5 \cos θ - 4) = 0 \Rightarrow \cos θ = 0, \cos θ = \frac{4}{5} Putting \cos θ = \frac{4}{5} in \sin θ + 2 \cos θ = 1, we get \Rightarrow \sin θ = 1 - 2 \cos θ = 1 - 2 (\frac{4}{5}) = - \frac{3}{5} This is not possible . Putting \cos θ = 0 in \sin θ + 2 \cos θ = 1, we get sinθ = 1 Thus, the value of 2 \sin θ - \cos θ = 2 (1) - 0 = 2 .$

Page No 531:

Question 42:

Prove the following trigonometric identities.

$If T_{n} = \sin^{n} θ + \cos^{n} θ, prove that \frac{T_{3} - T_{5}}{T_{1}} = \frac{T_{5} - T_{7}}{T_{3}}$

Answer:

In the given question, we are given

We need to prove

Here L.H.S is

Now, solving the L.H.S, we get

Further using the property, we get

So,

Now, solving the R.H.S, we get

So,

Further using the property, we get,

So,

Hence proved.

Page No 531:

Question 43:

Prove the following trigonometric identities.

If a cos³ θ + 3 a cos θ sin² θ = m, a sin³ θ + 3 a cos² θ sin θ = n, prove that (m + n)^2/3 + (m − n)^2/3 = 2a²^/3

Answer:

Given that,

We have to prove

Adding both the equations, we get

Also.

Therefore, we have

Hence proved.

Page No 531:

Question 44:

Prove the following trigonometric identities.

If x = a cos³ θ, y = b sin³ θ, prove that ${(\frac{x}{a})}^{2 / 3} + {(\frac{y}{b})}^{2 / 3} = 1$

Answer:

Given:

We have to prove

We know that

So, we have

Hence proved.

Page No 531:

Question 45:

Prove the following trigonometric identities.

If a cos θ + b sin θ = m and a sin θ − b cos θ = n, prove that a² + b² = m² + n²

Answer:

Given:

We have to prove

We know that,

Now, squaring and adding the two equations, we get

Hence proved.

Page No 531:

Question 46:

Prove the following trigonometric identities.

if cos A + cos²A = 1, prove that sin² A + sin⁴ A = 1

Answer:

Given:

We have to prove

Now,

Therefore, we have

Hence proved.

Page No 531:

Question 47:

If cos θ + cos² θ = 1, prove that sin¹² θ + 3 sin¹⁰ θ + 3 sin⁸ θ + sin⁶ θ + 2 sin⁴ θ + 2 sin² θ − 2 = 1

Answer:

Given:

We have to prove

From the given equation, we have

Therefore, we have

Hence proved.

Page No 531:

Question 48:

Given that:
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)

Show that one of the values of each member of this equality is sin α sin β sin γ

Answer:

Given:

Let us assume that

We know that,

Then, we have

Therefore, we have

Taking the expression with the positive sign, we have

Page No 531:

Question 49:

If sin θ + cos θ = x, prove that $\sin^{6} θ + \cos^{6} θ = \frac{4 - 3 {(x^{2} - 1)}^{2}}{4}$ .

Answer:

Given:

Squaring the given equation, we have

Squaring the last equation, we have

Therefore, we have

Hence proved.

Page No 531:

Question 50:

If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, show that $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1$ .

Answer:

Given:

We have to prove that .

Squaring the above equations and then subtracting the third from the sum of the first two, we have

Hence proved.

Page No 536:

Question 1:

If $\sin θ = \frac{1}{\sqrt{2}}$ , find all other trigonometric ratios of angle θ.

Answer:

Given:

We have to find all the trigonometric ratios.

We have the following right angle triangle.

From the above figure,

$Base = \sqrt{{Hypotenuse}^{2} - {Perpendicular}^{2}} \Rightarrow B C = \sqrt{A C^{2} - A B^{2}} \Rightarrow B C = \sqrt{{(\sqrt{2})}^{2} - 1^{2}} \Rightarrow B C = 1$

Page No 536:

Question 2:

If $\tan θ = \frac{1}{\sqrt{2}}$ , find the value of $\frac{{cosec}^{2} θ - \sec^{2} θ}{{cosec}^{2} θ + \cot^{2} θ}$ .

Answer:

Given:
We have to find the value of the expression

We know that,

Therefore, the given expression can be written as

Hence, the value of the given expression is .

Page No 536:

Question 3:

If 4tanθ = 3, evaluate $\frac{4 \sin θ - \cos θ + 1}{4 \sin θ + \cos θ - 1}$

Answer:

Given: 4tanθ = 3 ⇒ tan θ = $\frac{3}{4}$
Let us suppose a right angle triangle ABC right angled at B, with one of the acute angle θ. Let the sides be BC = 3k and AB = 4k, where k is a positive number.

By Pythagoras theorem, we get
${AC}^{2} = {BC}^{2} + {AB}^{2} {AC}^{2} = {(3 k)}^{2} + {(4 k)}^{2} {AC}^{2} = 9 k^{2} + 16 k^{2} AC = \sqrt{25 k^{2}} AC = \pm 5 k$
Ignoring AC = − 5k , as k is a positive number, we get
AC = 5k
If $\tan θ = \frac{BC}{AB} = \frac{3}{4}$ , then $\sin θ = \frac{BC}{AC} = \frac{3}{5}$ and $\cos θ = \frac{AB}{AC} = \frac{4}{5}$
Putting the values in $(\frac{4 \sin θ - \cos θ + 1}{4 \sin θ + \cos θ - 1})$ , we get
$(\frac{4 \times \frac{3}{5} - \frac{4}{5} + 1}{4 \times \frac{3}{5} + \frac{4}{5} - 1}) = (\frac{\frac{12 - 4 + 5}{5}}{\frac{12 + 4 - 5}{5}}) = \frac{13}{11}$

Page No 536:

Question 4:

If $\tan θ = \frac{12}{5}$ , find the value of $\frac{1 + \sin θ}{1 - \sin θ}$ .

Answer:

Given:
We have to find the value of the expression .

From the above figure, we have

Therefore,

Hence, the value of the given expression is 25.

Page No 536:

Question 5:

If $\cot θ = \frac{1}{\sqrt{3}}$ , find the value of $\frac{1 - \cos^{2} θ}{2 - \sin^{2} θ}$ .

Answer:

Given:
We have to find the value of the expression

We know that,

Using the identity , we have

Hence, the value of the given expression is $\frac{3}{5}$ .

Page No 536:

Question 6:

If $cosec A = \sqrt{2}$ , find the value of $\frac{2 \sin^{2} A + 3 \cot^{2} A}{4 (\tan^{2} A - \cos^{2} A)}$ .

Answer:

Given:
We have to find the value of the expression

We know that,

Therefore,

Hence, the value of the given expression is 2.

Page No 536:

Question 7:

If $\cot θ = \sqrt{3}$ , find the value of $\frac{\cos e c^{2} θ + \cot^{2} θ}{{cosec}^{2} θ - \sec^{2} θ}$ .

Answer:

Given:

We have to find the value of the expression .

We know that,

Therefore,

Hence, the value of the given expression is $\frac{21}{8}$ .

Page No 536:

Question 8:

If 3cosθ = 1, find the value of $\frac{6 \sin^{2} θ + \tan^{2} θ}{4 \cos θ}$

Answer:

Given:

We have to find the value of the expression .

We have,

Therefore,

Hence, the value of the expression is 10.

Page No 537:

Question 9:

If $\sqrt{3} \tan θ = 3 \sin θ$ , find the value of sin²θ − cos²θ.

Answer:

Given:

We have to find the value of .

Therefore,

Hence, the value of the expression is $\frac{1}{3}$ .

Page No 537:

Question 10:

If $cosec θ = \frac{13}{12}$ , find the value of $\frac{2 \sin θ - 3 \cos θ}{4 \sin θ - 9 \cos θ}$ .

Answer:

Given:

We have to find the value of the expression .

Now,

Therefore,

Hence, the value of the expression is 3.

Page No 537:

Question 11:

If sin θ + cos θ = $\sqrt{2} \cos (90 ° - θ)$ , find cot θ.

Answer:

Given:

We have to find the value of .

Now,

Hence,

Page No 537:

Question 12:

If $2 \sin^{2} θ - \cos^{2} θ = 2$ , then find the value of $θ$ .

Answer:

It is given that,
$2 \sin^{2} θ - \cos^{2} θ = 2 \Rightarrow 2 \sin^{2} θ - 2 = \cos^{2} θ \Rightarrow - 2 (1 - \sin^{2} θ) = \cos^{2} θ \Rightarrow - 2 \cos^{2} θ = \cos^{2} θ (1 - \sin^{2} θ = \cos^{2} θ) \Rightarrow 3 \cos^{2} θ = 0 \Rightarrow \cos θ = 0 \Rightarrow θ = 90 °$

Page No 537:

Question 13:

If $\sqrt{3} \tan θ - 1 = 0,$ find the value of sin²θ – cos²θ.

Answer:

Given: $\sqrt{3} \tan θ - 1 = 0,$
$\Rightarrow \tan θ = \frac{1}{\sqrt{3}} \Rightarrow \tan θ = \tan 30 ° \Rightarrow θ = 30 °$

$\begin{array}{rcl} Therefore, \sin^{2} θ - \cos^{2} θ & = & \sin^{2} (30 °) - \cos^{2} (30 °) \\ = & {(\frac{1}{2})}^{2} - {(\frac{\sqrt{3}}{2})}^{2} \\ = & \frac{1}{4} - \frac{3}{4} \\ = & \frac{- 2}{4} \\ = & \frac{- 1}{2} \end{array}$

Page No 537:

Question 1:

Define an identity.

Answer:

An identity is an equation which is true for all values of the variable (s).

For example,

Any number of variables may involve in an identity.

An example of an identity containing two variables is

The above are all about algebraic identities. Now, we define the trigonometric identities.

An equation involving trigonometric ratios of an angle (say) is said to be a trigonometric identity if it is satisfied for all valued of for which the trigonometric ratios are defined.

For examples,

$\sin^{2} θ + \cos^{2} θ = 1 1 + \tan^{2} θ = \sec^{2} θ 1 + \cot^{2} θ = {cosec}^{2} θ$

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Question 2:

What is the value of (1 − cos² θ) cosec² θ?

Answer:

We have,

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Question 3:

What is the value of (1 + cot² θ) sin² θ?

Answer:

We have,

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Question 4:

What is the value of $\sin^{2} θ + \frac{1}{1 + \tan^{2} θ}$ ?

Answer:

We have,

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Question 5:

Write the value of cosec² (90° − θ) − tan² θ.

Answer:

We have,

We know that,

Therefore, ${cosec}^{2} (90 ° - θ) - \tan^{2} θ = 1$

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Question 6:

Write the value of sin A cos (90° − A) + cos A sin (90° − A).

Answer:

We have,

We know that,

Therefore, $\sin A \cos (90 ° - A) + \cos A \sin (90 ° - A) = 1$

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Question 7:

Write the value of $\cot^{2} θ - \frac{1}{\sin^{2} θ}$ .

Answer:

We have,

We know that,

Therefore, $\cot^{2} θ - \frac{1}{\sin^{2} θ} = - 1$

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Question 8:

If x = a sin θ and y = b cos θ, what is the value of b²x² + a²y²?

Answer:

Given:

x = a sinθ and y = b cosθ

So,

$b^{2} x^{2} + a^{2} y^{2} = b^{2} {(a sinθ)}^{2} + a^{2} {(b cosθ)}^{2} = a^{2} b^{2} \sin^{2} θ + a^{2} b^{2} \cos^{2} θ = a^{2} b^{2} (\sin^{2} θ + \cos^{2} θ)$

We know that,

Therefore, $b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2}$

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Question 9:

If $\sin θ = \frac{4}{5}$ , what is the value of cotθ + cosecθ?

Answer:

Given:

We know that,

We have,

Hence, the value of cotθ + cosecθ is 2.

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Question 10:

What is the value of 9cot² θ − 9cosec² θ?

Answer:

We have,

We know that,

Therefore, $9 \cot^{2} θ - 9 {cosec}^{2} θ = - 9$

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Question 11:

What is the value of $6 \tan^{2} θ - \frac{6}{\cos^{2} θ}$ .

Answer:

We have,

We know that,

Therefore, $6 \tan^{2} θ - \frac{6}{\cos^{2} θ} = - 6$

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Question 12:

What is the value of $\frac{\tan^{2} θ - \sec^{2} θ}{\cot^{2} θ - {cosec}^{2} θ}$ .

Answer:

We have,

We know that,

Therefore,

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Question 13:

What is the value of (1 + tan² θ) (1 − sin θ) (1 + sin θ)?

Answer:

We have,

We know that,

Therefore,

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Question 14:

If $\cos A = \frac{7}{25}$ , find the value of tan A + cot A.

Answer:

Given:

We know that,

Therefore,

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Question 15:

If $\sin θ = \frac{1}{3}$ , then find the value of 2cot² θ + 2.

Answer:

Given:

We know that,

Therefore,

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Question 16:

If $\cos θ = \frac{3}{4}$ , then find the value of 9tan² θ + 9.

Answer:

Given:

We know that,

Therefore,

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Question 17:

If sec² θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.

Answer:

Given:

Hence, the value of k is 1.

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Question 18:

If cosec² θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.

Answer:

Given:

Thus, the value of λ is 1.

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Question 19:

If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.

Answer:

Given:

We know that,

Therefore,

Hence,

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Question 20:

If cosec θ − cot θ = α, write the value of cosec θ + cot α.

Answer:

Given:

We know that,

Therefore,

Hence,

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Question 21:

If sin² θ cos² θ (1 + tan² θ) (1 + cot² θ) = λ, then find the value of λ.

Answer:

Given:

$\Rightarrow λ = 1 \times 1 = 1$

Hence, the value of λ is 1.

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Question 22:

If 5x = sec θ and $\frac{5}{x} = \tan θ$ , find the value of $5 (x^{2} - \frac{1}{x^{2}})$ .

Answer:

Given:

We know that,

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Question 23:

If cosec θ = 2x and $\cot θ = \frac{2}{x}$ , find the value of $2 (x^{2} - \frac{1}{x^{2}})$ .

Answer:

Given:

We know that,

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Question 24:

Write True' or False' and justify your answer in each of the following :
(i) The value of $\sin θ$ is $x + \frac{1}{x}$ , where 'x' is a positive real number .
(ii) $\cos θ = \frac{a^{2} + b^{2}}{2 ab}$ , where a and b are two distinct numbers such that ab > 0.

(iii) The value of $\cos^{2} 23 - \sin^{2} 67$ is positive .

(iv) The value of the expression $\sin 80^{°} - \cos 80^{°}$ is negative .

(v) The value of $\sin θ + \cos θ$ is always greater than 1 .

Answer:

(i)
$\sin θ = x + \frac{1}{x} \Rightarrow - 1 \leq x + \frac{1}{x} \leq 1 \Rightarrow x + \frac{1}{x} \leq 1 \Rightarrow x^{2} + 1 \leq x \Rightarrow x^{2} + 1 - x \leq 0 Take x = 1, \Rightarrow 1 + 1 - 1 \leq 0 \Rightarrow 1 \leq 0 Which is false, so x is not always a positive real number . The given statement is false .$

(ii)
It is given that,
$\cos θ = \frac{a^{2} + b^{2}}{2 a b} Since, {(a - b)}^{2} = a^{2} + b^{2} - 2 a b > 0 [∵ {(a - b)}^{2} > 0, a and b are distint numbers] \Rightarrow a^{2} + b^{2} > 2 a b \Rightarrow \frac{a^{2} + b^{2}}{2 a b} > 1 \Rightarrow \cos θ > 1 But, - 1 \leq \cos θ \leq 1 Therefore, the given statement is false .$

(iii)
$\cos^{2} 23 ° - \sin^{2} 67 ° = \sin^{2} (90 ° - 23 °) - \sin^{2} 67 ° = \sin^{2} 67 ° - \sin^{2} 67 ° = 0$
Which is not positive, the given statement is false.

(iv)
Consider the table.

$θ$	0º	30º	45º	60º	90º
sin $θ$	0	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	1
cos $θ$	1	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	0

Here,

\sin 60 ° - \cos 60 ° = \frac{\sqrt{3}}{2} - \frac{1}{2} > 0 \sin 90 ° - \cos 90 ° = 1 - 0 > 0 So, \sin 80 ° - \cos 80 ° > 0 (\sin θ - \cos θ \geq 0 \forall 45 ° \leq θ \leq 90 °)

Therefore, the given statement is false.

(v)
Consider the table.

$θ$	0º	30º	45º	60º	90º
sin $θ$	0	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	1
cos $θ$	1	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	0

Here,

\sin 90 ° + \cos 90 ° = 1 + 0 = 1, which is not greater than 1 Therefore, the given statement is false .

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Question 25:

What is the value of cos² 67° – sin² 23°?

Answer:

$\cos^{2} 67 ° - \sin^{2} 23 ° = \cos^{2} 67 ° - \sin^{2} (90 ° - 67 °) = \cos^{2} 67 ° - \cos^{2} (67 °) (∵ \sin^{2} (90 - θ) = \cos^{2} θ) = 0$

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Question 1:

If $\frac{\cos 80 °}{\sin 10 °} + \cos 59 ° cosec 31 ° =____________.$

Answer:

$\frac{\cos 80 °}{\sin 10 °} + \cos 59 ° cosec 31 °$
$= \frac{\cos (90 ° - 10 °)}{\sin 10 °} + \cos 59 ° cosec (90 ° - 59 °) = \frac{\sin 10 °}{\sin 10 °} + \cos 59 ° \sec 59 ° = 1 + \cos 59 ° \times \frac{1}{\cos 59 °} = 1 + 1 = 2$

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Question 2:

The value of sin θ cos(90° – θ) + cos θ sin (90° – θ) is ___________.

Answer:

$\sin θ \cos (90 ° - θ) + \cos θ \sin (90 ° - θ) = \sin θ \times \sin θ + \cos θ \times \cos θ [\cos (90 ° - θ) = \sin θ and \sin (90 ° - θ) = \cos θ] = \sin^{2} θ + \cos^{2} θ = 1$

The value of sin θ cos(90° – θ) + cos θ sin (90° – θ) is _____1_____.

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Question 3:

The value of $\frac{\sin^{4} θ - \cos^{4} θ}{\sin^{2} θ - \cos^{2} θ}$ is ____________.

Answer:

$\frac{\sin^{4} θ - \cos^{4} θ}{\sin^{2} θ - \cos^{2} θ} = \frac{{(\sin^{2} θ)}^{2} - {(\cos^{2} θ)}^{2}}{\sin^{2} θ - \cos^{2} θ} = \frac{(\sin^{2} θ - \cos^{2} θ) (\sin^{2} θ + \cos^{2} θ)}{\sin^{2} θ - \cos^{2} θ} [a^{2} - b^{2} = (a - b) (a + b)]$
$= \sin^{2} θ + \cos^{2} θ = 1$

The value of $\frac{\sin^{4} θ - \cos^{4} θ}{\sin^{2} θ - \cos^{2} θ}$ is ______1______.

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Question 4:

\frac{1}{1 + \sin θ} + \frac{1}{1 - \sin θ} = k \sec^{2} θ

, then the value of k is _________.

Answer:

$\frac{1}{1 + \sin θ} + \frac{1}{1 - \sin θ} = \frac{1 - \sin θ + 1 + \sin θ}{(1 + \sin θ) (1 - \sin θ)} = \frac{2}{1 - \sin^{2} θ}$
$= \frac{2}{\cos^{2} θ} (\cos^{2} θ + \sin^{2} θ = 1) = 2 \sec^{2} θ$

Comparing with the given expression, we get

k = 2

If $\frac{1}{1 + \sin θ} + \frac{1}{1 - \sin θ} = k \sec^{2} θ$ , then the value of k is ____2____.

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Question 5:

If $\sqrt{(1 - \cos^{2} θ) \sec^{2} θ} = k$ tan θ and 0 < θ < 90°, then k = ________.

Answer:

$\sqrt{(1 - \cos^{2} θ) \sec^{2} θ} = \sqrt{\frac{\sin^{2} θ}{\cos^{2} θ}} (\sin^{2} θ + \cos^{2} θ = 1) = \sqrt{\tan^{2} θ} = \tan θ$

Comparing with the given expression, we get

k = 1

If $\sqrt{(1 - \cos^{2} θ) \sec^{2} θ} = k$ tan θ and 0 < θ < 90°, then k = ____1____.

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Question 6:

If cosec θ + cot θ = 3, then cos θ = _________.

Answer:

$cosec θ + \cot θ = 3 \Rightarrow \frac{1}{\sin θ} + \frac{\cos θ}{\sin θ} = 3 \Rightarrow \frac{1 + \cos θ}{\sin θ} = 3$
$\Rightarrow 1 + \cos θ = 3 \sin θ$
Squaring on both sides, we get
$1 + \cos^{2} θ + 2 \cos θ = 9 (1 - \cos^{2} θ) \Rightarrow 10 \cos^{2} θ + 2 \cos θ - 8 = 0 \Rightarrow 5 \cos^{2} θ + \cos θ - 4 = 0 \Rightarrow 5 \cos^{2} θ + 5 \cos θ - 4 \cos θ - 4 = 0$
$\Rightarrow 5 \cos θ (\cos θ + 1) - 4 (\cos θ + 1) = 0 \Rightarrow (\cos θ + 1) (5 \cos θ - 4) = 0 \Rightarrow \cos θ + 1 = 0 or 5 \cos θ - 4 = 0 \Rightarrow \cos θ = \frac{4}{5} (\cos θ cannot be negative in first quadrant)$

If cosec θ + cot θ = 3, then cos θ = $\frac{4}{5}$ .

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Question 7:

If cosec θ – cot θ = 2, then find the value of cosec² + cot²θ is ______.

Answer:

We know that,

${cosec}^{2} θ - \cot^{2} θ = 1 \Rightarrow (cosec θ - \cot θ) (cosec θ + \cot θ) = 1 \Rightarrow 2 (cosec θ + \cot θ) = 1 (Given) \Rightarrow cosec θ + \cot θ = \frac{1}{2}$
Now,
${(cosec θ - \cot θ)}^{2} + {(cosec θ + \cot θ)}^{2} = 2^{2} + {(\frac{1}{2})}^{2} \Rightarrow {cosec}^{2} θ + \cot^{2} θ - 2 cosec θ \cot θ + {cosec}^{2} θ + \cot^{2} θ + 2 cosec θ \cot θ = 4 + \frac{1}{4} \Rightarrow 2 ({cosec}^{2} θ + \cot^{2} θ) = \frac{17}{4} \Rightarrow {cosec}^{2} θ + \cot^{2} θ = \frac{17}{8}$

If cosec θ – cot θ = 2, then find the value of cosec²θ + cot²θ is $\frac{17}{8}$ .

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Question 8:

If sec θ + tan θ = m, then the value of sec⁴ θ – tan⁴ θ – 2 sec θ tan θ is _________.

Answer:

$\sec^{4} θ - \tan^{4} θ - 2 \sec θ \tan θ = (\sec^{2} θ - \tan^{2} θ) (\sec^{2} θ + \tan^{2} θ) - 2 \sec θ \tan θ = \sec^{2} θ + \tan^{2} θ - 2 \sec θ \tan θ (1 + \tan^{2} θ = \sec^{2} θ) = {(\sec θ - \tan θ)}^{2}$
$= {[\frac{(\sec θ + \tan θ) (\sec θ - \tan θ)}{\sec θ + \tan θ}]}^{2} = {(\frac{\sec^{2} θ - \tan^{2} θ}{\sec θ + \tan θ})}^{2} = {(\frac{1}{m})}^{2} = \frac{1}{m^{2}}$

If sec θ + tan θ = m, then the value of sec⁴ θ – tan⁴ θ – 2 sec θ tan θ is $\frac{1}{m^{2}}$ .

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Question 9:

The value of $\frac{\cos^{4} θ + \cos^{2} θ \sin^{2} θ + \sin^{2} θ}{\cos^{2} θ + \cos^{2} θ \sin^{2} θ + \sin^{4} θ}$ is __________.

Answer:

$\frac{\cos^{4} θ + \cos^{2} θ \sin^{2} θ + \sin^{2} θ}{\cos^{2} θ + \cos^{2} θ \sin^{2} θ + \sin^{4} θ} = \frac{{(\cos^{2} θ)}^{2} + \cos^{2} θ \sin^{2} θ + \sin^{2} θ}{\cos^{2} θ + \cos^{2} θ \sin^{2} θ + \sin^{4} θ} = \frac{{(1 - \sin^{2} θ)}^{2} + \cos^{2} θ \sin^{2} θ + \sin^{2} θ}{\cos^{2} θ + \cos^{2} θ \sin^{2} θ + \sin^{4} θ} = \frac{1 + \sin^{4} θ - 2 \sin^{2} θ + \cos^{2} θ \sin^{2} θ + \sin^{2} θ}{\cos^{2} θ + \cos^{2} θ \sin^{2} θ + \sin^{4} θ}$
$= \frac{1 - \sin^{2} θ + \sin^{4} θ + \cos^{2} θ \sin^{2} θ}{\cos^{2} θ + \cos^{2} θ \sin^{2} θ + \sin^{4} θ} = \frac{\cos^{2} θ + \cos^{2} θ \sin^{2} θ + \sin^{4} θ}{\cos^{2} θ + \cos^{2} θ \sin^{2} θ + \sin^{4} θ} = 1$

The value of $\frac{\cos^{4} θ + \cos^{2} {θsin}^{2} θ + \sin^{2} θ}{\cos^{2} θ + \cos^{2} {θsin}^{2} θ + \sin^{4} θ}$ is _____1_____.

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Question 10:

If sin θ – cos θ = $\frac{3}{5}$ , then sin θ cos θ = _______.

Answer:

$\sin θ - \cos θ = \frac{3}{5}$
Squaring on both sides, we get
${(\sin θ - \cos θ)}^{2} = {(\frac{3}{5})}^{2} \Rightarrow \sin^{2} θ + \cos^{2} θ - 2 \sin θ \cos θ = \frac{9}{25} \Rightarrow 1 - 2 \sin θ \cos θ = \frac{9}{25} \Rightarrow 2 \sin θ \cos θ = 1 - \frac{9}{25} = \frac{16}{25}$
$\Rightarrow \sin θ \cos θ = \frac{8}{25}$

If sin θ – cos θ = $\frac{3}{5}$ , then sin θ cos θ = $\frac{8}{25}$ .

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Question 11:

The value of sin²2° + sin²4° + sin²6° + ... + sin²90° is ____________.

Answer:

sin²2° + sin²4° + sin²6° + ... + sin²90°

= sin²2° + sin²4° + sin²6° + ... + sin²84° + sin²86° + sin²88°+ sin²90°

= [(sin²2° + sin²88°) + (sin²4° + sin²86°) + (sin²6° + sin²84°) + .... 22 pair of terms] + sin²90°

= [(sin²2° + cos²2°) + (sin²4° + cos²4°) + (sin²6° + cos²6°) + .... 22 pair of terms] + sin²90° [sin(90º − θ) = cosθ]

= (1 + 1 + 1 + ... 22 terms) + 1 [sin²θ + cos²θ = 1 and sin90° = 1]

= 22 + 1

= 23

The value of sin²2° + sin²4° + sin²6° + ... + sin²90° is ______23______.

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Question 12:

If $\frac{\sin^{2} θ - 3 sinθ + 2}{\cos^{2} θ} = 1, then θ =_______.$

Answer:

$\frac{\sin^{2} θ - 3 \sin θ + 2}{\cos^{2} θ} = 1 \Rightarrow \sin^{2} θ - 3 \sin θ + 2 = 1 - \sin^{2} θ \Rightarrow 2 \sin^{2} θ - 3 \sin θ + 1 = 0$
$\Rightarrow 2 \sin^{2} θ - 2 \sin θ - \sin θ + 1 = 0 \Rightarrow 2 \sin θ (\sin θ - 1) - 1 (\sin θ - 1) = 0 \Rightarrow (\sin θ - 1) (2 \sin θ - 1) = 0 \Rightarrow \sin θ - 1 = 0 or 2 \sin θ - 1 = 0$
$\Rightarrow \sin θ = 1 or \sin θ = \frac{1}{2} \Rightarrow θ = 90 ° or θ = 30 °$
Here, $θ$ cannot take the value 90º. For $θ = 90 °$ , the LHS of the given equation is not defined.

Thus, the value of $θ$ is $30 °$ .

If $\frac{\sin^{2} θ - 3 sinθ + 2}{\cos^{2} θ} = 1, then θ = 30 ° .$

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Question 13:

If cos θ + cos²θ = 1, then sin²θ + sin⁴θ = _________.

Answer:

$\cos θ + \cos^{2} θ = 1 \Rightarrow \cos θ = 1 - \cos^{2} θ \Rightarrow \cos θ = \sin^{2} θ$
Squaring on both sides, we get
$\cos^{2} θ = \sin^{4} θ \Rightarrow 1 - \sin^{2} θ = \sin^{4} θ \Rightarrow \sin^{2} θ + \sin^{4} θ = 1$

If cos θ + cos²θ = 1, then sin²θ + sin⁴θ = ____1_____.

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Question 14:

If $\frac{\tan^{3} θ - 1}{tanθ - 1} = A \sec^{2} θ + B \tan θ, then A + B =_______.$

Answer:

$\frac{\tan^{3} θ - 1}{\tan θ - 1} = \frac{(\tan θ - 1) (\tan^{2} θ + \tan θ + 1)}{\tan θ - 1} [a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})] = \tan^{2} θ + \tan θ + 1 = \sec^{2} θ + \tan θ (1 + \tan^{2} θ = \sec^{2} θ)$

Comparing with the given expression, we get

A = 1 and B = 1

∴ A + B = 2

If $\frac{\tan^{3} θ - 1}{tanθ - 1} = A \sec^{2} θ + B \tan θ, then A + B = 2 .$

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Question 15:

The value of (cosec θ – sin θ) (secθ – cos θ) (tan θ + cot θ) is _________.

Answer:

$(cosec θ - \sin θ) (\sec θ - \cos θ) (\tan θ + \cot θ) = (\frac{1}{\sin θ} - \sin θ) (\frac{1}{\cos θ} - \cos θ) (\frac{\sin θ}{\cos θ} + \frac{\cos θ}{\sin θ}) = (\frac{1 - \sin^{2} θ}{\sin θ}) (\frac{1 - \cos^{2} θ}{\cos θ}) (\frac{\sin^{2} θ + \cos^{2} θ}{\sin θ \cos θ})$
$= \frac{\cos^{2} θ}{\sin θ} \times \frac{\sin^{2} θ}{\cos θ} \times \frac{1}{\sin θ \cos θ} (\cos^{2} θ + \sin^{2} θ = 1) = 1$

The value of (cosec θ – sin θ) (secθ – cos θ) (tan θ + cot θ) is ___1___.

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Question 16:

$\sqrt{- 4 + \sqrt{8 + 16 {cosec}^{4} θ + \sin^{4} θ}} = A cosec θ + B \sin θ, then A =_______and B =_________.$

Answer:

$\sqrt{- 4 + \sqrt{8 + 16 {cosec}^{4} θ + \sin^{4} θ}} = \sqrt{- 4 + \sqrt{[{(4 {cosec}^{2} θ)}^{2} + 2 \times 4 {cosec}^{2} θ \times \sin^{2} θ + {(\sin^{2} θ)}^{2}]}} = \sqrt{- 4 + \sqrt{{(4 {cosec}^{2} θ + \sin^{2} θ)}^{2}}}$
$= \sqrt{- 4 + 4 {cosec}^{2} θ + \sin^{2} θ} = \sqrt{[{(2 cosec θ)}^{2} - 2 \times 2 cosec θ \times \sin θ + \sin^{2} θ]} = \sqrt{{(2 cosec θ - \sin θ)}^{2}} = 2 cosec θ - \sin θ$

Comparing with the given expression, we get

A = 2 and B = −1

$\sqrt{- 4 + \sqrt{8 + 16 {cosec}^{4} θ + \sin^{4} θ}} = A cosec θ + B \sin θ, then A = 2 and B = - 1 .$

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Question 17:

If (tan θ + 2) (2tan θ + 1) = A tan θ + B sec²θ, then AB = _________.

Answer:

$(\tan θ + 2) (2 \tan θ + 1) = 2 \tan^{2} θ + \tan θ + 4 \tan θ + 2 = 2 (\tan^{2} θ + 1) + 5 \tan θ = 2 \sec^{2} θ + 5 \tan θ (1 + \tan^{2} θ = \sec^{2} θ)$

Comparing with the given expression, we get

A = 5 and B = 2

∴ AB = 5 × 2 = 10

If (tan θ + 2) (2tan θ + 1) = A tan θ + B sec²θ, then AB = ____10____.

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Question 18:

If 2 sin θ + 3 cos θ = 2, then 3 sin θ – 2 cos θ = _________.

Answer:

${(2 \sin θ + 3 \cos θ)}^{2} + {(3 \sin θ - 2 \cos θ)}^{2} = 4 \sin^{2} θ + 9 \cos^{2} θ + 12 \sin θ \cos θ + 9 \sin^{2} θ + 4 \cos^{2} θ - 12 \sin θ \cos θ = 13 \sin^{2} θ + 13 \cos^{2} θ = 13 (\sin^{2} θ + \cos^{2} θ) = 13 (\sin^{2} θ + \cos^{2} θ = 1)$
$∴ {(2)}^{2} + {(3 \sin θ - 2 \cos θ)}^{2} = 13 (2 \sin θ + 3 \cos θ = 2) \Rightarrow {(3 \sin θ - 2 \cos θ)}^{2} = 13 - 4 = 9 \Rightarrow 3 \sin θ - 2 \cos θ = \pm 3$

If 2 sin θ + 3 cos θ = 2, then 3 sin θ – 2 cos θ = ____±3_____.

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Question 19:

If sin⁴ A – cos⁴ A = 1 and 0 < A ≤ 90°, then A = __________.

Answer:

$\sin^{4} A - \cos^{4} A = 1 \Rightarrow (\sin^{2} A - \cos^{2} A) (\sin^{2} A + \cos^{2} A) = 1 \Rightarrow \sin^{2} A - \cos^{2} A = 1 (\sin^{2} θ + \cos^{2} θ = 1) \Rightarrow 1 - \sin^{2} A + \cos^{2} A = 0$
$\Rightarrow 2 \cos^{2} A = 0 \Rightarrow \cos A = 0 \Rightarrow A = 90 °$

If sin⁴ A – cos⁴ A = 1 and 0 < A ≤ 90°, then A = $90 °$ .

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