NCERT Solutions for Class 12 Science Math Chapter 2 - Inverse Trigonometric Functions

Answer:

Let sin^-1 Then sin y =

We know that the range of the principal value branch of sin⁻¹ is

and sin

Therefore, the principal value of

Answer:

We know that the range of the principal value branch of cos⁻¹ is

.

Therefore, the principal value of.

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Let cosec⁻¹ (2) = y. Then,

We know that the range of the principal value branch of cosec⁻¹ is

Therefore, the principal value of

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We know that the range of the principal value branch of tan⁻¹ is

Therefore, the principal value of

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We know that the range of the principal value branch of cos⁻¹ is

Therefore, the principal value of

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Let tan⁻¹ (−1) = y. Then,

We know that the range of the principal value branch of tan⁻¹ is

Therefore, the principal value of

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We know that the range of the principal value branch of sec⁻¹ is

Therefore, the principal value of

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We know that the range of the principal value branch of cot⁻¹ is (0,π) and

Therefore, the principal value of

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We know that the range of the principal value branch of cos⁻¹ is [0,π] and

.

Therefore, the principal value of

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We know that the range of the principal value branch of cosec⁻¹ is

Therefore, the principal value of

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Answer:

Answer:

It is given that sin⁻¹ x = y.

We know that the range of the principal value branch of sin⁻¹ is

Therefore,.

Answer:

Answer:

To prove:

Let x = sinθ. Then,

We have,

R.H.S. =

= 3θ

= L.H.S.

Answer:

To prove:

Let x = cosθ. Then, cos⁻¹ x =θ.

We have,

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To prove:

Answer:

To prove:

Answer:

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Put x = cosec θ ⇒ θ = cosec⁻¹ x

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Answer:

$$\begin{gathered} {\tan }^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right) \\ ={\tan }^{-1}\left(\frac{1-{\displaystyle \frac{\sin x}{\cos x}}}{1+{\displaystyle \frac{\sin x}{\cos x}}}\right) \\ ={\tan }^{-1}\left(\frac{1-\tan x}{1+\tan x}\right) \\ ={\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(\tan x\right) \left({\tan }^{-1}\frac{x-y}{1+xy}={\tan }^{-1}x-{\tan }^{-1}y\right) \\ =\frac{\mathrm{\pi }}{4}-x \end{gathered}$$

Answer:

Answer:

Answer:

Let. Then,

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Answer:

Let x = tan θ. Then, θ = tan⁻¹ x.

Let y = tan Φ. Then, Φ = tan⁻¹ y.

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On squaring both sides, we get:

Hence, the value of x is

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Hence, the value of x is

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We know that sin⁻¹ (sin x) = x if, which is the principal value branch of sin⁻¹x.

Here,

Now, can be written as:

Answer:

We know that tan⁻¹ (tan x) = x if, which is the principal value branch of tan⁻¹x.

Here,

Now, can be written as:

Answer:

Let. Then,

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We know that cos⁻¹ (cos x) = x if, which is the principal value branch of cos ⁻¹x.

Here,

Now, can be written as:

$$\begin{gathered} {\cos }^{-1}\left(\cos \frac{7\mathrm{\pi }}{6}\right) = {\cos }^{-1}\left[\cos \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\right] \\ {\cos }^{-1}\left(\cos \frac{7\mathrm{\pi }}{6}\right) = {\cos }^{-1}\left[- \cos \frac{\mathrm{\pi }}{6}\right] \left[\mathrm{as}, \cos \left(\mathrm{\pi }+\mathrm{\theta }\right) = - \cos \mathrm{\theta }\right] \\ {\cos }^{-1}\left(\cos \frac{7\mathrm{\pi }}{6}\right) = {\cos }^{-1}\left[- \cos \left(\mathrm{\pi }-\frac{5\mathrm{\pi }}{6}\right)\right] \\ {\cos }^{-1}\left(\cos \frac{7\mathrm{\pi }}{6}\right) = {\cos }^{-1}\left[-\left\{- \cos \left(\frac{5\mathrm{\pi }}{6}\right)\right\}\right] \left[\mathrm{as}, \cos \left(\mathrm{\pi }-\mathrm{\theta }\right) = - \cos \mathrm{\theta }\right] \end{gathered}$$

The correct answer is B.

Answer:

Let. Then,

We know that the range of the principal value branch of.

∴

The correct answer is D.

Answer:

Let. Then,

We know that the range of the principal value branch of

Let.

The range of the principal value branch of

The correct answer is B.

Answer:

We know that cos⁻¹ (cos x) = x if, which is the principal value branch of cos ⁻¹x.

Here,

Now, can be written as:

Answer:

We know that tan⁻¹ (tan x) = x if, which is the principal value branch of tan ⁻¹x.

Here,

Now, can be written as:

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Now, we have:

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Now, we have:

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Now, we will prove that:

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Now, we have:

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Using (1) and (2), we have

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Answer:

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Answer:

Answer:

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Let tan⁻¹ x = y. Then,

The correct answer is D.

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Therefore, from equation (1), we have

Put x = sin y. Then, we have:

But, when, it can be observed that:

is not the solution of the given equation.

Thus, x = 0.

Hence, the correct answer is C.

Answer:

Hence, the correct answer is C.