Question

If $A={\cot }^{-1}\sqrt{\tan \theta }-{\tan }^{-1}\sqrt{\tan \theta }$, then $\tan (\frac{\pi }{4}-\frac{A}{2})$ is equal to

• A. $\sqrt{\cot \theta }$
• B. $\tan \theta$
• C. $\sqrt{\tan \theta }$
• D. None of these

Answer 1

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

Let $\sqrt{\tan \theta }=\tan \alpha$

$\therefore A={\cot }^{-1}\left(\tan \alpha \right)-{\tan }^{-1}\left(\tan \alpha \right)$

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

$=\frac{\pi }{2}-\alpha -\alpha =\frac{\pi }{2}-2\alpha$

$\Rightarrow 2\alpha =\frac{\pi }{2}-A\Rightarrow \alpha =\frac{\pi }{4}-\frac{A}{2}$

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