Question

If $u=co{t}^{-1}\sqrt{tan\alpha }-ta{n}^{-1}\sqrt{tan\alpha },thentan(\frac{\pi }{4}-\frac{u}{2})$ is equal to

• A. $\sqrt{tan\alpha }$
• B. $\sqrt{cot\alpha }$
• C. $tan\alpha$
• D. $cot\alpha$

Answer 1

The correct option is A $\sqrt{tan\alpha }$

We know that,

$ta{n}^{-1}\alpha +co{t}^{-1}\alpha =\frac{\pi }{2}$

$u=\frac{\pi }{2}-2{\tan }^{-1}\left(\sqrt{\tan \alpha }\right)$

$2{\tan }^{-1}\left(\sqrt{tan\alpha }\right)=\frac{\pi }{2}-u$

${\tan }^{-1}\left(\sqrt{\tan \alpha }\right)=\frac{\pi }{4}-\frac{u}{2}$

$\sqrt{tan\alpha }=\tan (\frac{\pi }{4}-\frac{u}{2})$