Rbse Solutions Class 12 Maths Exercise 2.1 (Principal Values)

Get complete NCERT Solutions for Class 12 Maths Chapter 2, Exercise 2.1 (Inverse Trigonometric Functions). Master finding the Principal Value of sin⁻¹, cos⁻¹, and tan⁻¹.

The Principal Value Branch (PVB) is crucial for these solutions. The PVB for each function is listed below:

Function	Principal Value Branch (PVB)
$\sin^{-1}(x)$	$[-\pi/2, \pi/2]$
$\cos^{-1}(x)$	$[0, \pi]$
$\tan^{-1}(x)$	$(-\pi/2, \pi/2)$
$\csc^{-1}(x)$	$[-\pi/2, \pi/2] – \{0\}$
$\sec^{-1}(x)$	$[0, \pi] – \{\pi/2\}$
$\cot^{-1}(x)$	$(0, \pi)$

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Question 1. $\sin^{-1}\left(-\frac{1}{2}\right)$

Solution:

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

We know $\sin(\pi/6) = 1/2$.

Since $\sin(-x) = -\sin x$, we have $\sin y = -\sin(\pi/6) = \sin(-\pi/6)$.

The principal value is $-\pi/6$, which lies in the PVB of $\sin^{-1}x$, i.e., $[-\pi/2, \pi/2]$.

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

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Question 2. $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Solution:

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

We know $\cos(\pi/6) = \frac{\sqrt{3}}{2}$.

The principal value is $\pi/6$, which lies in the PVB of $\cos^{-1}x$, i.e., $[0, \pi]$.

$$\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$$

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Question 3. $\csc^{-1}(2)$

Solution:

Let $y = \csc^{-1}(2)$. Then $\csc y = 2$.

We know $\csc(\pi/6) = 2$.

The principal value is $\pi/6$, which lies in the PVB of $\csc^{-1}x$, i.e., $[-\pi/2, \pi/2] – \{0\}$.

$$\csc^{-1}(2) = \frac{\pi}{6}$$

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Question 4. $\tan^{-1}(-\sqrt{3})$

Solution:

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

We know $\tan(\pi/3) = \sqrt{3}$.

Since $\tan(-x) = -\tan x$, we have $\tan y = -\tan(\pi/3) = \tan(-\pi/3)$.

The principal value is $-\pi/3$, which lies in the PVB of $\tan^{-1}x$, i.e., $(-\pi/2, \pi/2)$.

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

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Question 5. $\cos^{-1}\left(-\frac{1}{2}\right)$

Solution:

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

We know that for $\cos^{-1}(-x)$, we use the formula: $\cos^{-1}(-x) = \pi – \cos^{-1}(x)$.

$$\cos^{-1}\left(-\frac{1}{2}\right) = \pi – \cos^{-1}\left(\frac{1}{2}\right)$$

Since $\cos(\pi/3) = 1/2$, $\cos^{-1}(1/2) = \pi/3$.

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

The principal value $2\pi/3$ (which is $120^{\circ}$) lies in the PVB of $\cos^{-1}x$, i.e., $[0, \pi]$.

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

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Question 6. $\tan^{-1}(-1)$

Solution:

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

We know $\tan(\pi/4) = 1$.

Since $\tan(-x) = -\tan x$, $\tan y = -\tan(\pi/4) = \tan(-\pi/4)$.

The principal value is $-\pi/4$, which lies in the PVB of $\tan^{-1}x$, i.e., $(-\pi/2, \pi/2)$.

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

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Question 7. $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Solution:

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

We know $\sec(\pi/6) = \frac{2}{\sqrt{3}}$.

The principal value is $\pi/6$, which lies in the PVB of $\sec^{-1}x$, i.e., $[0, \pi] – \{\pi/2\}$.

$$\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6}$$

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Question 8. $\cot^{-1}(\sqrt{3})$

Solution:

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

We know $\cot(\pi/6) = \sqrt{3}$.

The principal value is $\pi/6$, which lies in the PVB of $\cot^{-1}x$, i.e., $(0, \pi)$.

$$\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}$$

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Question 9. $\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)$

Solution:

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

Using the property $\cos^{-1}(-x) = \pi – \cos^{-1}(x)$:

$$\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \pi – \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)$$

Since $\cos(\pi/4) = 1/\sqrt{2}$, $\cos^{-1}(1/\sqrt{2}) = \pi/4$.

$$\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \pi – \frac{\pi}{4} = \frac{4\pi – \pi}{4} = \frac{3\pi}{4}$$

The principal value $3\pi/4$ lies in the PVB of $\cos^{-1}x$, i.e., $[0, \pi]$.

$$\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4}$$

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Question 10. $\csc^{-1}(-\sqrt{2})$

Solution:

Let $y = \csc^{-1}(-\sqrt{2})$. Then $\csc y = -\sqrt{2}$.

We know $\csc(\pi/4) = \sqrt{2}$.

Since $\csc(-x) = -\csc x$, $\csc y = -\csc(\pi/4) = \csc(-\pi/4)$.

The principal value is $-\pi/4$, which lies in the PVB of $\csc^{-1}x$, i.e., $[-\pi/2, \pi/2] – \{0\}$.

$$\csc^{-1}(-\sqrt{2}) = -\frac{\pi}{4}$$

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Question 11. $\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)$

Solution:

We calculate the principal value of each term:

1. $\tan^{-1}(1)$: We found this is $\pi/4$.
2. $\cos^{-1}(-1/2)$: We found this is $2\pi/3$.
3. $\sin^{-1}(-1/2)$: We found this is $-\pi/6$.

Now, sum the values:

$$P = \frac{\pi}{4} + \frac{2\pi}{3} + \left(-\frac{\pi}{6}\right)$$

The Least Common Multiple (LCM) of 4, 3, and 6 is 12.

$$P = \frac{3\pi}{12} + \frac{8\pi}{12} – \frac{2\pi}{12}$$

$$P = \frac{(3 + 8 – 2)\pi}{12} = \frac{9\pi}{12}$$

Simplifying the fraction:

$$P = \frac{3\pi}{4}$$

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Question 12. $\cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)$

Solution:

We calculate the principal value of each term:

1. $\cos^{-1}(1/2)$: This value is $\pi/3$.
2. $\sin^{-1}(1/2)$: This value is $\pi/6$.

Now, substitute the values:

$$P = \cos^{-1}\left(\frac{1}{2}\right) + 2 \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} + 2 \left(\frac{\pi}{6}\right)$$

$$P = \frac{\pi}{3} + \frac{2\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3}$$

$$P = \frac{2\pi}{3}$$

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Question 13. If $\sin^{-1} x = y$, then:

(A) $0 \le y \le \pi$

(B) $-\pi/2 \le y \le \pi/2$

(C) $0 < y < \pi$

(D) $-\pi/2 < y < \pi/2$

Solution:

The question asks for the range or Principal Value Branch (PVB) of the function $f(x) = \sin^{-1}x$.

The PVB of $\sin^{-1}x$ is defined as the closed interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

This corresponds to the inequality $-\pi/2 \le y \le \pi/2$.

Correct Answer: (B)

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Question 14. $\tan^{-1}(\sqrt{3}) – \sec^{-1}(-2)$

Solution:

We calculate the principal value of each term:

1. $\tan^{-1}(\sqrt{3})$: This value is $\pi/3$.
2. $\sec^{-1}(-2)$: We use the property $\sec^{-1}(-x) = \pi – \sec^{-1}(x)$.$$\sec^{-1}(-2) = \pi – \sec^{-1}(2)$$Since $\sec(\pi/3) = 2$, $\sec^{-1}(2) = \pi/3$.$$\sec^{-1}(-2) = \pi – \frac{\pi}{3} = \frac{2\pi}{3}$$

Now, subtract the values:

$$P = \tan^{-1}(\sqrt{3}) – \sec^{-1}(-2) = \frac{\pi}{3} – \frac{2\pi}{3}$$

$$P = -\frac{\pi}{3}$$

Correct Answer: (B) $-\pi/3$

(Note: This question is multiple-choice in the textbook, with options typically being (A) $\pi$, (B) $-\pi/3$, (C) $\pi/3$, (D) $2\pi/3$. Based on the calculation, $-\pi/3$ is the correct result.)

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