Question

$sec{h}^{-1}(\sin \theta )$ is equal to

• A. $\log \tan \frac{\theta }{2}$
• B. $\log \sin \frac{\theta }{2}$
• C. $\log \cos \frac{\theta }{2}$
• D. $\log cot\frac{\theta }{2}$

Answer 1

The correct option is D

$\log cot\frac{\theta }{2}$

Explanation for the correct option:

Find the value of given expression:

As we know,

$\mathit{s}\mathit{e}\mathit{c}{\mathit{h}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{l}\mathit{o}\mathit{g}\left[\frac{1+\sqrt{(1–x^2)}}{x}\right]$

Substituting $x=\sin \theta$,

$\begin{array}{rcl}sec{h}^{-1}(sin\theta )& =& log\left[\frac{1+\sqrt{(1–\sin 2\theta })}{\sin \theta }\right]\\ & =& log\left[\frac{(1+\sqrt{\cos 2\theta })}{\sin \theta }\right]\\ & =& log\left[\frac{(1+\cos \theta )}{\sin \theta }\right]\\ & =& log\left[\frac{\left(1+{\cos }^2{\displaystyle \frac{\theta }{2}}–{\sin }^2{\displaystyle \frac{\theta }{2}}\right)}{2\sin {\displaystyle \frac{\theta }{2}}\cos {\displaystyle \frac{\theta }{2}}}\right]\\ & =& \log \left[\frac{\left({\cos }^2{\displaystyle \frac{\theta }{2}}+{\cos }^2{\displaystyle \frac{\theta }{2}}\right)}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\right]\\ & =& log\left[\frac{2{\cos }^2{\displaystyle \frac{\theta }{2}}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\right]\\ & =& \mathit{l}\mathit{o}\mathit{g}\left(cot\frac{\theta }{2}\right)\end{array}$ ….$$\begin{gathered} \left[\because cos\theta =co{s}^2\frac{\theta }{2}-si{n}^2\frac{\theta }{2}, \\ sin\theta =2sin\frac{\theta }{2}cos\frac{\theta }{2}, \\ si{n}^2\theta +co{s}^2\theta =1\right] \end{gathered}$$

Hence, Option ‘D’ is Correct.