Question

$co{t}^{-1}(\sqrt{\cos \alpha )}-{\tan }^{-1}(\sqrt{\cos \alpha )}=x$, then $\sin x$ is

• A. ${\tan }^2\left(\frac{\alpha }{2}\right)$
• B. $co{t}^2\frac{\alpha }{2}$
• C. $\tan \alpha$
• D. $cot\frac{\alpha }{2}$

Answer 1

The correct option is A

${\tan }^2\left(\frac{\alpha }{2}\right)$

Explanation for the correct option:

Given, $co{t}^{-1}\left(\sqrt{\cos \alpha }\right)-ta{n}^{-1}\left(\sqrt{\cos \alpha }\right)=x$

$\Rightarrow$ $ta{n}^{-1}\left(\frac{1}{\sqrt{\cos \alpha }}\right)-ta{n}^{-1}\left(\sqrt{\cos \alpha }\right)=x$

$\Rightarrow$ ${\tan }^{-1}\left[\frac{{\displaystyle \frac{1}{\sqrt{\cos \alpha }}}-\sqrt{\cos \alpha }}{1+\sqrt{{\displaystyle \frac{\cos \alpha }{\cos \alpha }}}}\right]=x$

$\Rightarrow$ ${\tan }^{-1}\left[\frac{{\displaystyle 1-\cos \alpha }}{2\sqrt{\cos \alpha }}\right]=x$

$\Rightarrow$ $\tan x=\frac{1-\cos \alpha }{2\sqrt{\cos \alpha }}$

Here, $P=1-\cos \alpha$ and $B=2\sqrt{\cos \alpha }$

$\Rightarrow$ $H=\sqrt{{(1-\cos \alpha )}^2+4\cos \alpha }$ ; $\left[\mathbf{\because }\mathit{H}\mathit{y}\mathit{p}\mathit{o}\mathit{t}\mathit{e}\mathit{n}\mathit{u}\mathit{s}\mathit{e}\mathbf{=}\sqrt{{\mathbf{Perpendicular}}^{\mathbf{2}}\mathbf{+}{\mathbf{Base}}^{\mathbf{2}}}\right]$

$=\sqrt{{(1+\cos \alpha )}^2}$

$=1+\cos \alpha$

$\therefore$$\sin x=\frac{P}{H}$

$=\frac{1-\cos \alpha }{1+\cos \alpha }$

$=\frac{2{\sin }^2{\displaystyle \frac{\alpha }{2}}}{2{\cos }^2{\displaystyle \frac{\alpha }{2}}}$

$=\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{2}}\frac{\mathbf{\alpha }\mathbf{}}{\mathbf{2}}$

Hence, Option ‘A’ is Correct.