Question

$\mathrm{If}u={\cot }^{-1}\sqrt{\tan \mathrm{\theta }}-{\tan }^{-1}\sqrt{\tan \mathrm{\theta }}\mathrm{then},\tan \left(\frac{\mathrm{\pi }}{4}-\frac{u}{2}\right)=$
(a) $\sqrt{\tan \mathrm{\theta }}$

(b) $\sqrt{\cot \mathrm{\theta }}$

(c) tan θ

(d) cot θ

Answer 1

(a) $\sqrt{\tan \mathrm{\theta }}$

Let $y=\sqrt{\tan \theta }$
Then,
$$\begin{gathered} u={\cot }^{-1}\sqrt{\tan \theta }-{\tan }^{-1}\sqrt{\tan \theta } \\ \Rightarrow u={\cot }^{-1}y-{\tan }^{-1}y \\ \Rightarrow u=\frac{\mathrm{\pi }}{2}-2{\tan }^{-1}y\left[\because {\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}\right] \\ \Rightarrow 2{\tan }^{-1}y=\frac{\mathrm{\pi }}{2}-u \\ \Rightarrow {\tan }^{-1}y=\frac{\mathrm{\pi }}{4}-\frac{u}{2} \\ \Rightarrow y=\tan \left(\frac{\mathrm{\pi }}{4}-\frac{u}{2}\right) \\ \Rightarrow \sqrt{\tan \theta }=\tan \left(\frac{\pi }{4}-\frac{u}{2}\right)\left[\because y=\sqrt{\tan \theta }\right] \end{gathered}$$