RD Sharma 2021 Solutions for Class 10 Maths Chapter 11 - Trigonometric Identities

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RD Sharma 2021 Solutions for Class 10 Maths Chapter 11 Trigonometric Identities are provided here with simple step-by-step explanations. These solutions for Trigonometric Identities are extremely popular among class 10 students for Maths Trigonometric Identities Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2021 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 RD Sharma 2021 Solutions. All RD Sharma 2021 Solutions for class 10 Maths are prepared by experts and are 100% accurate.

Page No 11.43:

Question 1:

Prove the following trigonometric identities.

(1 − cos² A) cosec² A = 1

Answer:

We know that,

So,

Page No 11.43:

Question 2:

Prove the following trigonometric identities.

(1 + cot² A) sin² A = 1

Answer:

We know that,

So,

Page No 11.43:

Question 3:

Prove the following trigonometric identities.

tan²θ cos²θ = 1 − cos²θ

Answer:

We know that, .

So,

Page No 11.43:

Question 4:

Prove the following trigonometric identities.

$cosec θ \sqrt{1 - \cos^{2} θ} = 1$

Answer:

We know that,

So,

Page No 11.43:

Question 5:

Prove the following trigonometric identities.

(sec² θ − 1) (cosec² θ − 1) = 1

Answer:

We know that,

So,

Page No 11.43:

Question 6:

Prove the following trigonometric identities.

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

Answer:

We know that,

So,

Page No 11.43:

Question 7:

Prove the following trigonometric identities.

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

Answer:

We know that,

Multiplying both numerator and the denominator by, we have

Page No 11.43:

Question 8:

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 11.43:

Question 9:

Prove the following trigonometric identities.

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

Answer:

We know that,

So,

Page No 11.43:

Question 10:

Prove the following trigonometric identities.

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

Answer:

We know that,

So,

Page No 11.43:

Question 11:

Prove the following trigonometric identities.

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

Answer:

We know that,

Multiplying numerator and denominator under the square root by , we have

Page No 11.43:

Question 12:

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 11.44:

Question 13:

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 11.44:

Question 14:

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 11.44:

Question 15:

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 11.44:

Question 16:

Prove the following trigonometric identities.

tan²θ − sin²θ = tan²θ sin²θ

Answer:

We have to prove

We know that,

So,

Page No 11.44:

Question 17:

Prove the following trigonometric identities.

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

Answer:

We have to prove

We know that,

So,

Page No 11.44:

Question 18:

Prove the following trigonometric identities.

(secθ + cosθ) (secθ − cosθ) = tan²θ + sin²θ

Answer:

We have to prove

We know that,

Page No 11.44:

Question 19:

Prove the following trigonometric identities.

secA (1 − sinA) (secA + tanA) = 1

Answer:

We have to prove

We know that,

So,

Page No 11.44:

Question 20:

Prove the following trigonometric identities.

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

Answer:

We have to prove

We know that,

So,

Page No 11.44:

Question 21:

Prove the following trigonometric identities.

(1 + tan²θ) (1 − sinθ) (1 + sinθ) = 1

Answer:

We have to prove

We know that,

So,

Page No 11.44:

Question 22:

Prove the following trigonometric identities.

sin²A cot²A + cos²A tan² A = 1

Answer:

We have to prove

We know that,

So,

Page No 11.44:

Question 23:

Prove the following trigonometric identities.

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

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

(iii) $\frac{\sin A - 2 \sin^{3} A}{2 \cos^{3} A - \cos A} = \tan A$

Answer:

(i) We have to prove

We know that,

So,

(ii) We have to prove

We know that,

So,

(iii)
Given
$\frac{sinA - 2 \sin^{3} A}{2 \cos^{3} A - cosA} = tanA$

$\begin{array}{rcl} LHS & = & \frac{sinA - 2 \sin^{3} A}{2 \cos^{3} A - cosA} \\ = & \frac{sinA (1 - 2 \sin^{2} A)}{cosA (2 \cos^{2} A - 1)} \\ = & tanA \frac{(\sin^{2} A + \cos^{2} A - 2 \sin^{2} A)}{(2 \cos^{2} A - \sin^{2} A - \cos^{2} A)} (∵ \sin^{2} A + \cos^{2} A = 1) \\ = & tanA (\frac{\cos^{2} A - \sin^{2} A}{\cos^{2} A - \sin^{2} A}) \\ = & tanA \\ = & RHS \end{array}$
Hence proved.

Page No 11.44:

Question 24:

Prove the following trigonometric identities.

$\frac{\cos^{2} θ}{\sin θ} - cosec θ + \sin θ = 0$

Answer:

We have to prove

We know that,

So,

Page No 11.44:

Question 25:

Prove the following trigonometric identities.

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

Answer:

We have to prove

We know that,

So,

Page No 11.44:

Question 26:

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 11.44:

Question 27:

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 11.44:

Question 28:

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 11.44:

Question 29:

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 11.44:

Question 30:

Prove the following trigonometric identities.

$\frac{\tan θ}{1 - \cot θ} + \frac{\cot θ}{1 - \tan θ} = 1 + \tan θ + \cot θ$

Answer:

We need to prove

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

Further using the identity, we get

Hence

Page No 11.44:

Question 31:

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 11.44:

Question 32:

Prove the following trigonometric identities

cosec⁶θ = cot⁶θ + 3 cot²θ cosec²θ + 1

Answer:

We need to prove

Solving the L.H.S, we get

()

Further using the identity, we get

Hence proved.

Page No 11.44:

Question 33:

Prove the following trigonometric identities.

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

Answer:

We need to prove

Solving the L.H.S, we get

Using , we get

Hence proved.

Page No 11.44:

Question 34:

Prove the following trigonometric identities.

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

Answer:

We need to prove

Using the property , we get

LHS =

Further using the identity, , we get

= RHS

Hence proved.

Page No 11.44:

Question 35:

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 11.44:

Question 36:

Prove the following trigonometric identities.

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

Answer:

We need to prove

Now, multiplying the numerator and denominator of LHS by , we get

Further using the identity, , we get

Hence proved.

Page No 11.44:

Question 37:

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 11.44:

Question 38:

Prove that:

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

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

(iii) $\sqrt{\frac{1 + \cos θ}{1 - \cos θ}} + \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} = 2 cosec θ$

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

Answer:

(i) We have,

â€‹

(ii) We have,

(iii) We have,

(iv) We have,

Multiplying both the numerator and the denominator by, we have

Page No 11.45:

Question 39:

Prove the following trigonometric identities.

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

Answer:

We need to prove

Here, we will first solve the L.H.S.

Now, using , we get

Further using the property , we get

So,

Hence proved.

Page No 11.45:

Question 40:

Prove the following trigonometric identities.

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

Answer:

We need to prove

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

Using and, we get

Hence proved.

Page No 11.45:

Question 41:

Prove the following trigonometric identities.

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

Answer:

We need to prove

Solving the L.H.S, we get

Further using the property , we get

So,

Hence proved.

Page No 11.45:

Question 42:

Prove the following trigonometric identities.

$\frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \sin A + \cos A$

Answer:

We need to prove

Solving the L.H.S, we get

= RHS

Hence proved.

Page No 11.45:

Question 43:

Prove the following trigonometric identities.

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

Answer:

We need to prove

Using the identity, we get

Further, using the property , we get

So,

Hence proved.

Page No 11.45:

Question 44:

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 11.45:

Question 45:

Prove the following trigonometric identities.

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

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,

Now, taking common from both the numerator and the denominator, we get

Hence proved.

Page No 11.45:

Question 46:

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 11.45:

Question 47:

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 11.45:

Question 48:

Prove the following trigonometric identities.

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

Answer:

In the given question, we need to prove .

Here, we will first solve the L.H.S.

Now, using, we get

On further solving, we get

Similarly we solve the R.H.S.

Now, using, we get

On further solving, we get

So, L.H.S = R.H.S

Hence proved.

Page No 11.45:

Question 49:

Prove the following trigonometric identities.

tan²A + cot²A = sec²A cosec²A − 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 11.45:

Question 50:

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

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 11.45:

Question 51:

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 11.45:

Question 52:

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 11.45:

Question 53:

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 11.45:

Question 54:

Prove the following trigonometric identities.

sin²A cos²B − cos²A sin²B = sin²A − sin²B

Answer:

We know that,

So have,

Hence proved.

Page No 11.45:

Question 55:

Prove the following trigonometric identities.

(i) $\frac{\cot A + \tan B}{\cot B + \tan A} = \cot A \tan B$

(ii) $\frac{\tan A + \tan B}{\cot A + \cot B} = \tan A \tan B$

Answer:

(i) We have to prove

Now,

Hence proved.

(ii) We have to prove

Now,

Hence proved.

Page No 11.45:

Question 56:

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 11.45:

Question 57:

Prove the following trigonometric identities.

tan² A sec² B − sec² A tan² B = tan² A − tan²B

Answer:

We have to prove

We know that,

So,

Hence proved.

Page No 11.45:

Question 58:

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 11.45:

Question 59:

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 11.46:

Question 60:

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 11.46:

Question 61:

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 11.46:

Question 62:

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 11.46:

Question 63:

Prove the following trigonometric identities.

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

Answer:

In the given question, we need to prove

Now, using the identity in L.H.S, we get

Further using, we get

Also, from the identity , we get

Hence proved.

Page No 11.46:

Question 64:

Prove the following trigonometric identities.

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

Answer:

In the given question, we need to prove

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

Further using the identity, we get

Further using the identity , we get

Now, from the identity, we get

So,

Hence proved.

Page No 11.46:

Question 65:

Prove the following trigonometric identities.

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

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.

(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 11.46:

Question 66:

Prove the following trigonometric identities.

(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A

Answer:

We have to prove

We know that,

So, we have

Hence proved.

Page No 11.46:

Question 67:

Prove the following trigonometric identities.

(1 + cot A − cosec A) (1 + tan A + sec A) = 2

Answer:

We have to prove

We know that, .

So,

Hence proved.

Page No 11.46:

Question 68:

Prove the following trigonometric identities.

(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)

Answer:

We have to prove

We know that,

Consider the LHS.

Now, consider the RHS.

∴ LHS = RHS

Hence proved.

Page No 11.46:

Question 69:

Prove the following trigonometric identities.

(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A

Answer:

We have to prove

We know that,

So,

Hence proved.

Page No 11.46:

Question 70:

Prove the following trigonometric identities.

(i) $\frac{\cos A cosec A - \sin A \sec A}{\cos A + \sin A} = cosec A - \sec A$

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

Answer:

(i) We have to prove

So,

Hence proved.

(ii) We have to prove

We know that,

So,

Hence proved.

Page No 11.46:

Question 71:

Prove the following trigonometric identities.

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

Answer:

We have to prove

We know that,

So,

Hence proved.

Page No 11.46:

Question 72:

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 11.46:

Question 73:

Prove the following trigonometric identities.

sec⁴ A(1 − sin⁴ A) − 2 tan² A = 1

Answer:

We have to prove

We know that,

So,

Hence proved.

Page No 11.46:

Question 74:

Prove the following trigonometric identities.

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

Answer:

We have to prove .

We know that,

So,

Multiplying both the numerator and denominator by , we have

Hence proved.

Page No 11.46:

Question 75:

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 11.46:

Question 76:

Prove the following trigonometric identities.

If $\frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1 and \frac{x}{a} \sin θ - \frac{y}{b} \cos θ = 1, prove that \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 2$

Answer:

Given that,

We have to prove

We know that,

Squaring and then adding the above two equations, we have

Page No 11.46:

Question 77:

Prove the following trigonometric identities.

If cosec θ − sin θ = a³, sec θ − cos θ = b³, prove that a²b² (a² + b²) = 1

Answer:

Given that,

We have to prove

We know that

Now from the first equation, we have

Again from the second equation, we have

Therefore, we have

Hence proved.

Page No 11.47:

Question 78:

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 11.47:

Question 79:

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 11.47:

Question 80:

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 11.47:

Question 81:

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 11.47:

Question 82:

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 11.47:

Question 83:

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 11.47:

Question 84:

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 11.47:

Question 85:

If x = a sec θ cos Ï•, y = b sec θ sin Ï• 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 11.47:

Question 86:

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 11.54:

Question 1:

If $\cos θ = \frac{4}{5}$ , find all other trigonometric ratios of angle θ.

Answer:

Given:

Now, we have to find all the other trigonometric ratios.

We have the following right angle triangle.

From the above figure,

$Perpendicular = \sqrt{{Hypotenuse}^{2} - {Base}^{2}}$

Therefore,

Page No 11.54:

Question 2:

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 11.54:

Question 3:

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 11.54:

Question 4:

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 11.54:

Question 5:

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 11.54:

Question 6:

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 11.54:

Question 7:

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 11.54:

Question 8:

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 11.54:

Question 9:

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 11.54:

Question 10:

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 11.54:

Question 11:

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 11.54:

Question 12:

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 11.54:

Question 13:

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 11.54:

Question 14:

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 11.55:

Question 1:

If sec θ + tan θ = x, then sec θ =

(a) $\frac{x^{2} + 1}{x}$
(b) $\frac{x^{2} + 1}{2 x}$
(c) $\frac{x^{2} - 1}{2 x}$
(d) $\frac{x^{2} - 1}{x}$

Answer:

Given:

We know that,

Now,

Adding the two equations, we get

Therefore, the correct choice is (b).

Page No 11.55:

Question 2:

If $secθ + tanθ = x$ , then $tanθ =$

(a) $\frac{x^{2} + 1}{x}$
(b) $\frac{x^{2} - 1}{x}$
(c) $\frac{x^{2} + 1}{2 x}$
(d) $\frac{x^{2} - 1}{2 x}$

Answer:

Given:

We know that,

Now,

Subtracting the second equation from the first equation, we get

Therefore, the correct choice is (d).

Page No 11.55:

Question 3:

$\sqrt{\frac{1 + \sin θ}{1 - \sin θ}}$ is equal to

(a) sec θ + tan θ
(b) sec θ − tan θ
(c) sec² θ + tan² θ
(d) sec² θ − tan² θ

Answer:

The given expression is .

Multiplying both the numerator and denominator under the root by , we have

Therefore, the correct option is (a).

Page No 11.55:

Question 4:

The value of $\sqrt{\frac{1 + \cos θ}{1 - \cos θ}}$ is

(a) cot θ − cosec θ
(b) cosec θ + cot θ
(c) cosec² θ + cot² θ
(d) (cot θ + cosec θ)²

Answer:

The given expression is .

Multiplying both the numerator and denominator under the root by, we have

Therefore, the correct choice is (b).

Page No 11.55:

Question 5:

sec⁴ A − sec² A is equal to

(a) tan² A − tan⁴ A
(b) tan⁴ A − tan² A
(c) tan⁴ A + tan² A
(d) tan² A + tan⁴ A

Answer:

The given expression is .

Taking common from both the terms, we have

Disclaimer: The options given in (c) and (d) are same by the commutative property of addition.

Therefore, the correct options are (c) or (d).

Page No 11.55:

Question 6:

cos⁴ A − sin⁴ A is equal to

(a) 2 cos² A + 1
(b) 2 cos² A − 1
(c) 2 sin² A − 1
(d) 2 sin² A + 1

Answer:

The given expression is .

Factorising the given expression, we have

Therefore, the correct option is (b).

Page No 11.55:

Question 7:

$\frac{\sin θ}{1 + \cos θ}$ is equal to

(a) $\frac{1 + \cos θ}{\sin θ}$
(b) $\frac{1 - \cos θ}{\cos θ}$
(c) $\frac{1 - \cos θ}{\sin θ}$
(d) $\frac{1 - \sin θ}{\cos θ}$

Answer:

The given expression is .

Multiplying both the numerator and denominator under the root by , we have

Therefore, the correct option is (c).

Page No 11.55:

Question 8:

$\frac{\sin θ}{1 - \cot θ} + \frac{\cos θ}{1 - \tan θ}$ is equal to

(a) 0
(b) 1
(c) sin θ + cos θ
(d) sin θ − cos θ

Answer:

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (c).

Page No 11.55:

Question 9:

The value of (1 + cot θ − cosec θ) (1 + tan θ + sec θ) is

(a) 1
(b) 2
(c) 4
(d) 0

Answer:

The given expression is

Simplifying the given expression, we have

Therefore, the correct option is (b).

Page No 11.55:

Question 10:

$\frac{\tan θ}{\sec θ - 1} + \frac{\tan θ}{\sec θ + 1}$ is equal to

(a) 2 tan θ
(b) 2 sec θ
(c) 2 cosec θ
(d) 2 tan θ sec θ

Answer:

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (c).

Page No 11.55:

Question 11:

(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to

(a) 0
(b) 1
(c) −1
(d) None of these

Answer:

The given expression is

Simplifying the given expression, we have

Therefore, the correct option is (b).

Page No 11.55:

Question 12:

If x = a cos θ and y = b sin θ, then b²x² + a²y² =

(a) a²b²
(b) ab
(c) a⁴b⁴
(d) a² + b²

Answer:

Given:

So,

We know that,

Therefore,

Hence, the correct option is (a).

Page No 11.56:

Question 13:

If x = a sec θ and y = b tan θ, then b²x² − a²y² =

(a) ab
(b) a² − b²
(c) a² + b²
(d) a²b²

Answer:

Given:

So,

We know that,

Therefore,

Hence, the correct option is (d).

Page No 11.56:

Question 14:

$\frac{\cot θ}{\cot θ - c o t 3 θ} + \frac{\tan θ}{\tan θ - \tan 3 θ}$ is equal to

(a) 0
(b) 1
(c) −1
(d) 2

Answer:

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (b).

Page No 11.56:

Question 15:

2 (sin⁶ θ + cos⁶ θ) − 3 (sin⁴ θ + cos⁴ θ) is equal to

(a) 0
(b) 1
(c) −1
(d) None of these

Answer:

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (c).

Page No 11.56:

Question 16:

If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a² + b² =

(a) 7
(b) 12
(c) 25
(d) None of these

Answer:

Given:

Squaring and then adding the above two equations, we have

Hence, the correct option is (c).

Page No 11.56:

Question 17:

If a cot θ + b cosec θ = p and b cot θ − a cosec θ = q, then p² − q² =

(a) a²− b²
(b) b² − a²
(c) a² + b²
(d) b − a

Answer:

Given:

Squaring both the equations and then subtracting the second from the first, we have

Hence, the correct option is (b).

Page No 11.56:

Question 18:

The value of sin² 29° + sin² 61° is

(a) 1
(b) 0
(c) 2 sin² 29°
(d) 2 cos² 61°

Answer:

The given expression is .

Hence, the correct option is (a).

Page No 11.56:

Question 19:

If x = r sin θ cos Ï•, y = r sin θ sin Ï• and z = r cos θ, then

(a) $x^{2} + y^{2} + z^{2} = r^{2}$
(b) $x^{2} + y^{2} - z^{2} = r^{2}$
(c) $x^{2} - y^{2} + z^{2} = r^{2}$
(d) $z^{2} + y^{2} - x^{2} = r^{2}$

Answer:

Given:

Squaring and adding these equations, we get

Hence, the correct option is (a).

Page No 11.56:

Question 20:

If sin θ + sin² θ = 1, then cos² θ + cos⁴ θ =

(a) −1
(b) 1
(c) 0
(d) None of these

Answer:

Given:

Now,

Hence, the correct option is (b).

Page No 11.56:

Question 21:

If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a² + b² =

(a) m² − n²
(b) m²n²
(c) n² − m²
(d) m² + n²

Answer:

Given:

Squaring and adding these equations, we have

Hence, the correct option is (d).

Page No 11.56:

Question 22:

If cos A + cos² A = 1, then sin² A + sin⁴ A =

(a) −1
(b) 0
(c) 1
(d) None of these

Answer:

Given:

So,

Hence, the correct option is (c).

Page No 11.56:

Question 23:

If x = a sec θ cos Ï•, y = b sec θ sin Ï• and z = c tan θ, then $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ =

(a) $\frac{z^{2}}{c^{2}}$
(b) $1 - \frac{z^{2}}{c^{2}}$
(c) $\frac{z^{2}}{c^{2}} - 1$
(d) $1 + \frac{z^{2}}{c^{2}}$

Answer:

Given:

Now,

Hence, the correct option is (d).

Page No 11.56:

Question 24:

If a cos θ − b sin θ = c, then a sin θ + b cos θ =

(a) $\pm \sqrt{a^{2} + b^{2} + c^{2}}$
(b) $\pm \sqrt{a^{2} + b^{2} - c^{2}}$
(c) $\pm \sqrt{c^{2} - a^{2} - b^{2}}$
(d) None of these

Answer:

Given:

Squaring on both sides, we have

Hence, the correct option is (b).

Page No 11.56:

Question 25:

9 sec² A − 9 tan² A is equal to

(a) 1
(b) 9
(c) 8
(d) 0

Answer:

Given:

We know that,

Therefore,

Hence, the correct option is (b).

Page No 11.57:

Question 26:

(1 + tan θ + sec θ) (1 + cot θ − cosec θ) =

(a) 0
(b) 1
(c) 1
(d) −1

Answer:

The given expression is .

Simplifying the given expression, we have

Disclaimer: None of the given options match with the answer.

Page No 11.57:

Question 27:

(sec A + tan A) (1 − sin A) =

(a) sec A
(b) sin A
(c) cosec A
(d) cos A

Answer:

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (d).

Page No 11.57:

Question 28:

$\frac{1 + \tan^{2} A}{1 + \cot^{2} A}$ is equal to

(a) sec²A
(b) −1
(c) cot²A
(d) tan²A

Answer:

Given:

Therefore, the correct option is (d).

Page No 11.57:

Question 29:

If sin $θ - \cos θ = 0$ , then the value of $\sin^{4} θ + \cos^{4} θ$ is
(a) 1 (b) $- 1$ (c) $\frac{1}{2}$ (d) $\frac{1}{4}$

Answer:

It is given that,
$\sin θ - \cos θ = 0 \Rightarrow \sin θ = \cos θ \Rightarrow \frac{\sin θ}{\cos θ} = 1 \Rightarrow \tan θ = 1 \Rightarrow \tan θ = \tan 45 ° \Rightarrow θ = 45 °$
$∴ \sin^{4} θ + \cos^{4} θ = \sin^{4} 45 ° + \cos^{4} 45 ° = {(\frac{1}{\sqrt{2}})}^{4} + {(\frac{1}{\sqrt{2}})}^{4} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
Hence, the correct answer is option (c).

Page No 11.57:

Question 30:

The value of sin ( $45^{°} + θ) - \cos (45^{°} - θ)$ is equal to
(a) 2 cos $θ$ (b) 0 (c) 2 sin $θ$ (d) 1

Answer:

We know that, $\sin (90 - θ) = \cos θ$ .
So, $\sin (45 ° + θ) = \cos [90 - (45 ° + θ)] = \cos (45 ° - θ)$
$∴ \sin (45 ° + θ) - \cos (45 ° - θ) = \cos (45 ° - θ) - \cos (45 ° - θ) = 0$
Hence, the correct answer is option (b).

Page No 11.57:

Question 31:

If $∆ A B C$ is right angled at C , then the value of cos ( $A + B$ ) is

Answer:

In a right angled triangle ABC, $∠ C$ is a right angle.
We know that, the sum of angles of a triangle is 180º.
$∴ ∠ A + ∠ B + ∠ C = 180 ° \Rightarrow ∠ A + ∠ B + 90 ° = 180 ° \Rightarrow ∠ A + ∠ B = 90 °$
$∴ \cos (A + B) = \cos 90 ° = 0$
Hence, the correct answer is option (a).

Page No 11.57:

Question 32:

If cos $9 θ$ = sin $θ$ and $9 θ$ < 90⁰ , then the value of tan $6 θ$ is

Answer:

It is given that,
$\cos 9 θ = \sin θ, 9 θ < 90 ° \Rightarrow \sin (90 ° - 9 θ) = \sin θ [\sin (90 ° - θ) = \cos θ] \Rightarrow 90 ° - 9 θ = θ \Rightarrow 10 θ = 90 ° \Rightarrow θ = 9 ° Therefore, \tan 6 θ = \tan 54 ° .$
Disclaimer: Answer of the given question is not matching with the options provided in the textbook.

Page No 11.57:

Question 33:

If cos ( $α + β$ ) = 0 , then sin $(α - β)$ can be reduced to

Answer:

It is given that,
$\cos (α + β) = 0 \Rightarrow \cos (α + β) = \cos 90 ° (\cos 90 ° = 0) \Rightarrow α + β = 90 ° \Rightarrow α = 90 ° - β Now, put α = 90 ° - β in \sin (α - β) . ∴ \sin (α - β) = \sin (90 ° - β - β) = \sin (90 ° - 2 β) = \cos 2 β [\sin (90 ° - θ) = \cos θ]$
Hence, the correct answer is option (b).

Page No 11.57:

Question 1:

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 2:

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 3:

The value of sin² 20° + sin² 70° is __________.

Answer:

We know that,
$\sin 70 ° = \sin (90 ° - 20 °) = \cos 20 ° [\sin (90 ° - θ) = \cos θ]$
$∴ \sin^{2} 20 ° + \sin^{2} 70 ° = \sin^{2} 20 ° + \cos^{2} 20 ° = 1 (\sin^{2} θ + \cos^{2} θ = 1)$

The value of sin² 20° + sin² 70° is _____1_____.

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Question 4:

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 5:

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 6:

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 7:

If sin⁴A – cos⁴A = 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 sin⁴A – cos⁴A = 1 and 0 < A ≤ 90°, then A = $90 °$ .

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Question 8:

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 9:

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 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:

\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 17:

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

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 18:

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 19:

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 20:

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 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,

Page No 11.58:

Question 3:

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

Answer:

We have,

Page No 11.58:

Question 4:

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

Answer:

We have,

Page No 11.59:

Question 5:

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,

Page No 11.59:

Question 6:

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 7:

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 8:

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 9:

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 10:

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 11:

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 12:

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 13:

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

Answer:

We have,

We know that,

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

Page No 11.59:

Question 14:

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,

Page No 11.59:

Question 15:

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

Answer:

We have,

We know that,

Therefore,

Page No 11.59:

Question 16:

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 17:

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

Answer:

Given:

We know that,

Therefore,

Page No 11.59:

Question 18:

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

Answer:

Given:

We know that,

Therefore,

Page No 11.59:

Question 19:

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.

Page No 11.59:

Question 20:

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 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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