Question

If $u={\cot }^{-1}\left\{\sqrt{\tan \theta }\right\}-{\tan }^{-1}\left\{\sqrt{\tan \theta }\right\}$ then find the value of: $\tan (\frac{\pi }{4}-\frac{u}{2})$

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

Answer 1

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

Given, $u={\cot }^{-1}\left(\sqrt{\tan \theta }\right)-{\tan }^{-1}\left(\sqrt{\tan \theta }\right)$

We know that,

${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$

$\Rightarrow {\cot }^{-1}x=\frac{\pi }{2}-{\tan }^{-1}x$

$\Rightarrow {\cot }^{-1}\sqrt{\tan \theta }=\frac{\pi }{2}-{\tan }^{-1}\sqrt{\tan \theta }$

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

$\Rightarrow u=\frac{\pi }{2}-2{\tan }^{-1}\sqrt{\tan \theta }$

$\Rightarrow \frac{u}{2}=\frac{\pi }{4}-{\tan }^{-1}\sqrt{\tan \theta }$

$\Rightarrow \frac{\pi }{4}-\frac{u}{2}={\tan }^{-1}\sqrt{\tan \theta }$

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

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