Question

If ${\tan }^{-1}(cosec{\tan }^{-1}x-\tan {\cot }^{-1}x)=a{\tan }^{-1}x(x\ne 0)$.
Find the value of a

• A. $a=1/2$
• B. $a=1$
• C. $a=2$
• D. $a=4$

Answer 1

The correct option is A $a=1/2$

We have, $2{\tan }^{-1}(cosec{\tan }^{-1}x-\tan {\cot }^{-1}x)=2a{\tan }^{-1}x(x\ne 0)$

Case (i):
$x<0$, put $x=-\left|x\right|$
L.H.S. $=2{\tan }^{-1}(cosec\left({\tan }^{-1}(-\left|x\right|)\right)-\tan \left({\cot }^{-1}(-\left|x\right|)\right))$

$=2{\tan }^{-1}(cosec(-{\tan }^{-1}\left|x\right|)-\tan (\pi -{\cot }^{-1}\left|x\right|))$

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

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

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

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

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

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

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

$=2{\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=2\times \frac{\theta }{2}=\theta ={\tan }^{-1}\left|x\right|$

Case (ii):
$x>0$

L.H.S. $=2{\tan }^{-1}(cosec\left({\tan }^{-1}x\right)-\tan \left({\cot }^{-1}x\right))$

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

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

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

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

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

$=2{\tan }^{-1}\left(\frac{2\sin \theta /2}{2\sin \theta /2\cos \theta /2}\right)=2{\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\theta ={\tan }^{-1}x$

Hence $a=1/2$