Question

2tan-1 (cosec tan-1 x - tan cot-1 x)=tan-1 x

Answer 1

that's the answer hope it helps you

Answer 2

Answer:

LHS=RHS proved below.

Step-by-step explanation:

Given : $2\tan^{-1}(\csc(\tan^{-1}x)- \tan(\cot^{-1}x))=\tan^{-1}x$

To find : To prove the above expression?

Solution :

Taking LHS,

$2\tan^{-1}(\csc(\tan^{-1}x)- \tan(\cot^{-1}x))$

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

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

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

Put $x=\tan \theta$

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

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

$=2\tan^{-1}(\frac{1-\cos\theta}{\sin\theta}$

$=2\tan^{-1}(\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}$

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

$=\theta$

$=\tan^{-1}x$

$=RHS$

LHS=RHS