Vardaan Learning Institute

Ans: Domain: $[-1, 1]$, Range: $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Ans: Domain: $[-1, 1]$, Range: $[0, \pi]$

Ans: Range: $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. It is an open interval, unlike $\sin^{-1}x$ which is closed.

Ans: Range: $[0, \pi] - \left\{\frac{\pi}{2}\right\}$. The angle $\frac{\pi}{2}$ is excluded because $\sec\left(\frac{\pi}{2}\right)$ is not defined.

Ans: True.

Ans: False. $\sin^{-1}x$ is the inverse sine function, while $(\sin x)^{-1} = \frac{1}{\sin x} = \text{cosec } x$.

Ans: $\frac{\pi}{6}$ (since $\sin(\frac{\pi}{6}) = \frac{1}{2}$ and $\frac{\pi}{6} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$)

Ans: $\frac{\pi}{6}$

Ans: $\frac{\pi}{4}$

Ans: $-\frac{\pi}{6}$ (Use $\sin^{-1}(-x) = -\sin^{-1}(x)$)

Ans: $\frac{2\pi}{3}$ (Use $\cos^{-1}(-x) = \pi - \cos^{-1}(x) = \pi - \frac{\pi}{3}$)

Ans: $-\frac{\pi}{3}$ (Use $\tan^{-1}(-x) = -\tan^{-1}(x)$)

Ans: $\frac{1}{4}$ (Since $\frac{1}{4} \in [-1, 1]$)

Ans: $\frac{\pi}{6}$ (Since $\frac{\pi}{6} \in [0, \pi]$)

Ans: $\sin^{-1}\left(\frac{1}{x}\right)$ (for $|x| \ge 1$)

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

Ans: $-\sin^{-1}(x)$

Ans: $\pi - \cos^{-1}(x)$

Ans: $-\tan^{-1}(x)$

Ans: $\frac{\pi}{2}$

Ans: $\frac{\pi}{2}$

Ans: $\frac{\pi}{2}$

Ans: $\tan^{-1}\left(\frac{x + y}{1 - xy}\right)$

Ans: $\tan^{-1}\left(\frac{x - y}{1 + xy}\right)$

Ans: $\tan^{-1}\left(\frac{2x}{1 - x^2}\right)$

Ans: $\sin^{-1}\left(\frac{2x}{1 + x^2}\right)$

Ans: $\cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right)$

Ans: $x = a \sin \theta$ or $x = a \cos \theta$

Ans: $x = a \tan \theta$ or $x = a \cot \theta$

Ans: $x = a \sec \theta$ or $x = a \text{cosec } \theta$