Question

$co{t}^{-1}\left[{(\cos \alpha )}^{½}\right]-{\tan }^{-1}\left[{(\cos \alpha )}^{½}\right]=x$, then $\sin x=$

• A. ${\tan }^2\left(\frac{\alpha }{2}\right)$
• B. $co{t}^2\left(\frac{\alpha }{2}\right)$
• 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, $co{t}^{-1}\left(\sqrt{\cos \alpha }\right)-{\tan }^{-1}\left(\sqrt{\cos \alpha }\right)=x$

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

$\Rightarrow$ ${\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{\sqrt{\cos \alpha }}-\sqrt{\cos \alpha }}}{1+{\displaystyle \frac{\sqrt{\cos \alpha }}{\sqrt{\cos \alpha }}}}\right)=x$ ; $\left[\mathbf{\because }\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{A}\mathbf{-}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{B}\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{A-B}{1+A.B}\right)\right]$

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

Here, $P=1-\cos \alpha ,B=2\cos \alpha$

$\Rightarrow$ $H=\sqrt{{\left(1-\cos \alpha \right)}^2+4\cos \alpha }$ ; $\left[\mathbf{\because }\mathit{H}\mathit{y}\mathit{p}\mathit{o}\mathit{t}\mathit{e}\mathit{n}\mathit{u}\mathit{s}\mathit{e}\mathbf{}\mathbf{=}\mathbf{}\sqrt{\mathbf{P}\mathbf{e}\mathbf{r}\mathbf{p}\mathbf{e}\mathbf{n}\mathbf{d}\mathbf{i}\mathbf{c}\mathbf{u}\mathbf{l}\mathbf{a}{\mathbf{r}}^{\mathbf{2}}\mathbf{+}\mathbf{B}\mathbf{a}\mathbf{s}{\mathbf{e}}^{\mathbf{2}}}\right]$

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

$=1+\cos \alpha$

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

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

$=\frac{2{\sin }^2{\displaystyle \frac{\alpha }{2}}}{2{\cos }^2{\displaystyle \frac{\alpha }{2}}}$ ; $\left[\mathbf{\because }\mathbf{1}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\frac{\mathbf{\theta }}{\mathbf{2}}\mathbf{,}\mathbf{}\mathbf{1}\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\frac{\mathbf{\theta }}{\mathbf{2}}\right]$

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

Hence, Option ‘A’ is Correct.