Class 12-NCERT Solutions-Chapter-02-Inverse Trigonometric Functions-Ex 2.1

Solution:

Let \( \sin^{-1} \left( -\frac{1}{2} \right) = y \). Then, \( \sin y = -\frac{1}{2} \).

We know that the range of the principal value branch of \( \sin^{-1} \) is \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).

\[ \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \]

\[ \Rightarrow -\sin \left( \frac{\pi}{6} \right) = -\frac{1}{2} \]

Using the property \( \sin(-x) = -\sin x \):

\[ \sin \left( -\frac{\pi}{6} \right) = -\frac{1}{2} \]

Since \( -\frac{\pi}{6} \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \), the principal value is \( -\frac{\pi}{6} \).

Therefore, the principal value of \( \sin^{-1} \left( -\frac{1}{2} \right) \) is \( -\frac{\pi}{6} \).

Solution:

Let \( \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) = y \). Then, \( \cos y = \frac{\sqrt{3}}{2} \).

We know that the range of the principal value branch of \( \cos^{-1} \) is \( [0, \pi] \).

\[ \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \]

Since \( \frac{\pi}{6} \in [0, \pi] \), the principal value is \( \frac{\pi}{6} \).

Therefore, the principal value of \( \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \) is \( \frac{\pi}{6} \).

Solution:

Let \( \text{cosec}^{-1} (2) = y \). Then, \( \text{cosec } y = 2 \).

We know that the range of the principal value branch of \( \text{cosec}^{-1} \) is \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\} \).

\[ \text{cosec } \left( \frac{\pi}{6} \right) = 2 \]

Since \( \frac{\pi}{6} \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\} \), the principal value is \( \frac{\pi}{6} \).

Therefore, the principal value of \( \text{cosec}^{-1} (2) \) is \( \frac{\pi}{6} \).

Solution:

Let \( \tan^{-1} (-\sqrt{3}) = y \). Then, \( \tan y = -\sqrt{3} \).

We know that the range of the principal value branch of \( \tan^{-1} \) is \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).

\[ \tan \left( \frac{\pi}{3} \right) = \sqrt{3} \]

Using the property \( \tan(-x) = -\tan x \):

\[ \Rightarrow \tan \left( -\frac{\pi}{3} \right) = -\sqrt{3} \]

Since \( -\frac{\pi}{3} \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \), the principal value is \( -\frac{\pi}{3} \).

Therefore, the principal value of \( \tan^{-1} (-\sqrt{3}) \) is \( -\frac{\pi}{3} \).

Solution:

Let \( \cos^{-1} \left( -\frac{1}{2} \right) = y \). Then, \( \cos y = -\frac{1}{2} \).

We know that the range of the principal value branch of \( \cos^{-1} \) is \( [0, \pi] \).

\[ \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \]

Using the property \( \cos(\pi - x) = -\cos x \):

\[ \begin{aligned} \cos \left( \pi - \frac{\pi}{3} \right) &= -\frac{1}{2} \\ \cos \left( \frac{2\pi}{3} \right) &= -\frac{1}{2} \end{aligned} \]

Since \( \frac{2\pi}{3} \in [0, \pi] \), the principal value is \( \frac{2\pi}{3} \).

Therefore, the principal value of \( \cos^{-1} \left( -\frac{1}{2} \right) \) is \( \frac{2\pi}{3} \).

Solution:

Let \( \tan^{-1} (-1) = y \). Then, \( \tan y = -1 \).

We know that the range of the principal value branch of \( \tan^{-1} \) is \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).

\[ \tan \left( \frac{\pi}{4} \right) = 1 \]

Using \( \tan(-x) = -\tan x \):

\[ \tan \left( -\frac{\pi}{4} \right) = -1 \]

Therefore, the principal value of \( \tan^{-1} (-1) \) is \( -\frac{\pi}{4} \).

Solution:

Let \( \sec^{-1} \left( \frac{2}{\sqrt{3}} \right) = y \). Then, \( \sec y = \frac{2}{\sqrt{3}} \).

The range of the principal value branch of \( \sec^{-1} \) is \( [0, \pi] - \left\{ \frac{\pi}{2} \right\} \).

\[ \sec \left( \frac{\pi}{6} \right) = \frac{2}{\sqrt{3}} \]

Since \( \frac{\pi}{6} \in [0, \pi] - \left\{ \frac{\pi}{2} \right\} \), the principal value is \( \frac{\pi}{6} \).

Therefore, the principal value is \( \frac{\pi}{6} \).

Solution:

Let \( \cot^{-1} (\sqrt{3}) = y \). Then, \( \cot y = \sqrt{3} \).

The range of the principal value branch of \( \cot^{-1} \) is \( (0, \pi) \).

\[ \cot \left( \frac{\pi}{6} \right) = \sqrt{3} \]

Since \( \frac{\pi}{6} \in (0, \pi) \), the principal value is \( \frac{\pi}{6} \).

Therefore, the principal value is \( \frac{\pi}{6} \).

Solution:

Let \( \cos^{-1} \left( -\frac{1}{\sqrt{2}} \right) = y \). Then, \( \cos y = -\frac{1}{\sqrt{2}} \).

The range of the principal value branch of \( \cos^{-1} \) is \( [0, \pi] \).

\[ \cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \]

Using \( \cos(\pi - x) = -\cos x \):

\[ \begin{aligned} \cos \left( \pi - \frac{\pi}{4} \right) &= -\frac{1}{\sqrt{2}} \\ \cos \left( \frac{3\pi}{4} \right) &= -\frac{1}{\sqrt{2}} \end{aligned} \]

Since \( \frac{3\pi}{4} \in [0, \pi] \), the principal value is \( \frac{3\pi}{4} \).

Therefore, the principal value is \( \frac{3\pi}{4} \).

Solution:

Let \( \text{cosec}^{-1} (-\sqrt{2}) = y \). Then, \( \text{cosec } y = -\sqrt{2} \).

The range of the principal value branch of \( \text{cosec}^{-1} \) is \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\} \).

\[ \text{cosec } \left( \frac{\pi}{4} \right) = \sqrt{2} \]

Using \( \text{cosec}(-x) = -\text{cosec } x \):

\[ \text{cosec } \left( -\frac{\pi}{4} \right) = -\sqrt{2} \]

Since \( -\frac{\pi}{4} \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] - \{0\} \), the principal value is \( -\frac{\pi}{4} \).

Therefore, the principal value is \( -\frac{\pi}{4} \).

Solution:

Let us find the principal values individually:

1. \( \tan^{-1}(1) = \frac{\pi}{4} \)

2. \( \cos^{-1} \left( -\frac{1}{2} \right) = \pi - \cos^{-1} \left( \frac{1}{2} \right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \)

3. \( \sin^{-1} \left( -\frac{1}{2} \right) = -\sin^{-1} \left( \frac{1}{2} \right) = -\frac{\pi}{6} \)

Now, substitute these values into the expression:

\[ \begin{aligned} &= \frac{\pi}{4} + \frac{2\pi}{3} + \left( -\frac{\pi}{6} \right) \\ &= \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} \\ &= \frac{3\pi + 8\pi - 2\pi}{12} \quad (\text{LCM of 4, 3, 6 is 12}) \\ &= \frac{9\pi}{12} \\ &= \frac{3\pi}{4} \end{aligned} \]

Therefore, the value of the expression is \( \frac{3\pi}{4} \).

Solution:

1. \( \cos^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{3} \)

2. \( \sin^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{6} \)

Substitute these values into the expression:

\[ \begin{aligned} &= \frac{\pi}{3} + 2 \left( \frac{\pi}{6} \right) \\ &= \frac{\pi}{3} + \frac{\pi}{3} \\ &= \frac{2\pi}{3} \end{aligned} \]

Therefore, the value of the expression is \( \frac{2\pi}{3} \).

Answer: (B)

Solution:

Given \( \sin^{-1} x = y \).

We know that the range of the principal value branch of \( \sin^{-1} \) is \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).

Therefore, \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \).

Answer: (B)

Solution:

1. \( \tan^{-1} \sqrt{3} = \frac{\pi}{3} \)

2. For \( \sec^{-1} (-2) \): Let \( \sec^{-1} (-2) = y \Rightarrow \sec y = -2 \). We know \( \sec \left( \frac{\pi}{3} \right) = 2 \). Using \( \sec(\pi - x) = -\sec x \): \( \sec \left( \pi - \frac{\pi}{3} \right) = -2 \Rightarrow \sec \left( \frac{2\pi}{3} \right) = -2 \). So, \( \sec^{-1} (-2) = \frac{2\pi}{3} \).

Now, substitute these values:

\[ \begin{aligned} &= \frac{\pi}{3} - \frac{2\pi}{3} \\ &= -\frac{\pi}{3} \end{aligned} \]

Therefore, the correct answer is (B).