Question

The value of $2{\tan }^{-1}(cosec{\tan }^{-1}x-\tan {\cot }^{-1}x)$ is equal to

• A. ${\cot }^{-1}x$
• B. ${\cot }^{-1}\frac{1}{x}$
• C. ${\tan }^{-1}x$
• D. $noneofthese$

Answer 1

Answer: (C)

${\tan }^{-1}x$

Given, $2{\tan }^{-1}(cosec{\tan }^{-1}x-\tan {\cot }^{-1}x)$

$⟹2{\tan }^{-1}(cosec\left({\cot }^{-1}\left(\frac{1}{x}\right)\right)-\tan \left({\tan }^{-1}\left(\frac{1}{x}\right)\right))$

$=2{\tan }^{-1}(\sqrt{1+{\cot }^2\left({\cot }^{-1}\left(\frac{1}{x}\right)\right)}-\frac{1}{x})$ ,,,,,,,,,,,,,,,, ${cosec}^{2}x=1+{\cot }^{2}x$

$=2{\tan }^{-1}(\sqrt{1+\frac{1}{x^2}}-\frac{1}{x})$

$=2{\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$

Let $x=\tan A$

Hence the above expression reduces, to

$=2{\tan }^{-1}\left(\frac{\sqrt{1+\tan A^2}-1}{\tan A}\right)$

$=2{\tan }^{-1}\left(\frac{\sec A+1}{\tan A}\right)$ ....... ${\sec }^2\theta =1+{\tan }^2\theta$

$=2{\tan }^{-1}\left(\frac{1+\frac{1}{cosA}}{\frac{sinA}{cosA}}\right)$ ......... $\sec x=\frac{1}{\cos x}$ and $\frac{\sin \theta }{\cos \theta }=\tan \theta$

$=2{\tan }^{-1}\left(\frac{1+\cos A}{\sin A}\right)$

$=2{\tan }^{-1}\left(\tan \frac{A}{2}\right)$

$=2\times \frac{A}{2}$

$=A$

$={\tan }^{-1}\left(x\right)$