Question

$2ta{n}^{-1}(cosecta{n}^{-1}x-tanco{t}^{-1}x)$ is equal to

• A. $si{n}^{-1}x$
• B. $co{s}^{-1}x$
• C. $ta{n}^{-1}x$
• D. $co{t}^{-1}x$

Answer 1

Answer: (C)

$ta{n}^{-1}x$

Let
$I=2ta{n}^{-1}(cosecta{n}^{-1}x-tanco{t}^{-1}x)=2ta{n}^{-1}(coseccose{c}^{-1}\left(\frac{\sqrt{1+x^2}}{x}\right)-tanta{n}^{-1}\left(\frac{1}{x}\right))=2ta{n}^{-1}(\frac{\sqrt{1+x^2}}{x}-\frac{1}{x})=2ta{n}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$
Substituting $x=tan\theta$, we get
$I=2ta{n}^{-1}\frac{sec\theta -1}{tan\theta }=2ta{n}^{-1}\left(tan\frac{\theta }{2}\right)=\theta ={tan}^{-1}x$