NCERT Solutions

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

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

Since $\sin(\frac{\pi}{6}) = \frac{1}{2}$, we have:

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

As $-\frac{\pi}{6} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, the principal value is $-\frac{\pi}{6}$.

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

Range of principal value branch of $\cos^{-1}$ is $[0, \pi]$.

Since $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\frac{\pi}{6} \in [0, \pi]$.

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

Range of $\text{cosec}^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}] – \{0\}$.

Since $\text{cosec}(\frac{\pi}{6}) = 2$.

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

Range is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Since $\tan(\frac{\pi}{3}) = \sqrt{3}$, we have $\tan(-\frac{\pi}{3}) = -\tan(\frac{\pi}{3}) = -\sqrt{3}$.

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

Range is $[0, \pi]$. Since the value is negative, it lies in the 2nd quadrant.

We know $\cos(\frac{\pi}{3}) = \frac{1}{2}$. Using $\cos(\pi – \theta) = -\cos \theta$:

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

Range is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

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

Range is $[0, \pi] – \{\frac{\pi}{2}\}$.

$\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \implies \sec(\frac{\pi}{6}) = \frac{2}{\sqrt{3}}$.

Range is $(0, \pi)$.

$\cot(\frac{\pi}{6}) = \sqrt{3}$.

Let $y = \cos^{-1}(-\frac{1}{\sqrt{2}})$. Range is $[0, \pi]$.

We know $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.

Since the value is negative, we use the property $\cos(\pi – \theta) = -\cos \theta$.

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

Range is $[-\frac{\pi}{2}, \frac{\pi}{2}] – \{0\}$.

$\text{cosec}(\frac{\pi}{4}) = \sqrt{2} \implies \text{cosec}(-\frac{\pi}{4}) = -\sqrt{2}$.

We evaluate each term separately:

• $\tan^{-1}(1) = \frac{\pi}{4}$
• $\cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3}$ (from Q5)
• $\sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}$ (from Q1)

Adding them up:

$$ \frac{\pi}{4} + \frac{2\pi}{3} + (-\frac{\pi}{6}) $$ $$ = \frac{3\pi + 8\pi – 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4} $$

• $\cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$
• $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$

Expression: $\frac{\pi}{3} + 2(\frac{\pi}{6}) = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}$.

The question asks for the Principal Value Branch (Range) of $\sin^{-1} x$.

By definition, the range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Therefore, $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$.

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

2. For $\sec^{-1}(-2)$: Let $y = \sec^{-1}(-2)$. Range is $[0, \pi]$.

$\sec y = -2$. Since $\sec(\frac{\pi}{3}) = 2$, we use $\sec(\pi – \theta) = -\sec \theta$.

$\sec(\pi – \frac{\pi}{3}) = -2 \implies y = \frac{2\pi}{3}$.

Expression: $\frac{\pi}{3} – \frac{2\pi}{3} = -\frac{\pi}{3}$.