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Q.
If $2 \, \tan^{-1} (\cos \, x) = \tan^{-1} (2 \,cosec \,x ), $ then the value of x is
Inverse Trigonometric Functions
Solution:
We know that
$2 \tan^{-1} x = \tan^{-1} \frac{2x}{1 - x^{2}} $
Hence: $2 \tan^{-1} \left(\cos x\right) = \tan^{-1} \left(\frac{2 \cos x}{1 - \cos^{2} x }\right)$
$ 2 \tan^{-1} \left(\cos x\right) = \tan^{-1} \left(2 cosec x\right) $
so, $\tan^{-1} \left(\frac{2 \cos x}{1- \cos^{2} x}\right) = \tan^{- 1} \left(2 cosec x \right) $
$\Rightarrow \frac{2 \cos x}{\sin^{2} x} = 2 cosec x$
$\Rightarrow \, 2 \cos x = 2 \sin^{2} x cosec x = 2 \sin x $ or $\tan x = 1$
so, $ x =\frac{\pi}{4}$ [The other value $ \frac{3\pi}{4} $ is not valid since this lies in the third quadrant]