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  4. tan((π/4)+(1/2) cos-1x)+ tan ((π/4)-(1/2) cos-1x),x ≠ 0 is equal to

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Q. $\tan\left(\frac{\pi}{4}+\frac{1}{2}\cos^{-1}x\right)+ tan \left(\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x\right),x \neq 0$ is equal to

Inverse Trigonometric Functions

Solution:

Let $cos^{-1} x= \theta$. Then the given expression becomes
$\frac{ 1+tan \frac{\theta}{2} }{1-tan \frac{\theta}{2}} +\frac{ 1- tan \frac{\theta}{2}}{1+tan \frac{\theta}{2}} $
$= \frac{2\left(1+tan ^{2} \frac{\theta}{2}\right)}{1-tan^{2} \frac{\theta}{2}}$
$ = \frac{2}{ cos\, \theta}$
$ = \frac{2}{x}$