Question

Write the value of the expression $\tan \left(\frac{{\sin }^{-1}x+{\cos }^{-1}x}{2}\right)$, where $x=\frac{\sqrt{3}}{2}$.

Answer 1

Given, $x=\frac{\sqrt{3}}{2}$

$\therefore \tan \left(\frac{{\sin }^{-1}x+{\cos }^{-1}x}{2}\right)$$=\tan \left(\frac{{\sin }^{-1}\frac{\sqrt{3}}{2}+{\cos }^{-1}\frac{\sqrt{3}}{2}}{2}\right)$

We know that,

$\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$ and $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$

$=\tan \left(\frac{\frac{\pi }{3}+\frac{\pi }{6}}{2}\right)$

$=\tan \left(\frac{\frac{2\pi }{6}+\frac{\pi }{6}}{2}\right)$

$=\tan \left(\frac{\frac{2\pi +\pi }{6}}{2}\right)$

$=\tan \left(\frac{\frac{3\pi }{6}}{2}\right)$

$=\tan \left(\frac{\frac{\pi }{2}}{2}\right)$

$=\tan \left(\frac{\pi }{4}\right)$

$=1$