Question

2 tan⁻¹ {cosec (tan⁻¹ x) − tan (cot⁻¹ x)} is equal to
(a) cot⁻¹ x

(b) cot⁻¹$\frac{1}{x}$

(c) tan⁻¹ x

(d) none of these

Answer 1

(c) tan⁻¹ x

Let ${\tan }^{-1}x=y$
So, $x=\tan y$

$$\begin{gathered} \therefore 2{\tan }^{-1}\left\{\mathrm{cosec}\left({\tan }^{-1}x\right)-\tan \left(co{t}^{-1}x\right)\right\}=2{\tan }^{-1}\left\{\mathrm{cosec}\left({\tan }^{-1}x\right)-\tan \left({\tan }^{-1}\frac{1}{x}\right)\right\} \\ =2{\tan }^{-1}\left\{\mathrm{cosec}\left({\tan }^{-1}x\right)-\frac{1}{x}\right\} \\ =2{\tan }^{-1}\left\{\mathrm{cosec}y-\frac{1}{\tan y}\right\} \\ =2{\tan }^{-1}\left\{\frac{1-\cos y}{\sin y}\right\} \\ =2{\tan }^{-1}\left\{\frac{2{\sin }^2{\displaystyle \frac{y}{2}}}{\sin y}\right\} \\ =2{\tan }^{-1}\left\{\frac{2{\sin }^2{\displaystyle \frac{y}{2}}}{2\sin {\displaystyle \frac{y}{2}}\cos {\displaystyle \frac{y}{2}}}\right\} \\ =2{\tan }^{-1}\left\{\tan \frac{y}{2}\right\} \\ =y \\ ={\tan }^{-1}x \end{gathered}$$