Question

Find m if the following equation holds true

$\tan \left(\frac{1}{2}{\sin }^{-1}\frac{3}{4}\right)=\frac{4-\sqrt{m}}{3}$

Answer 1

Let, $\tan \left(\frac{1}{2}{\sin }^{-1}\frac{3}{4}\right)=x$

$\frac{1}{2}{\sin }^{-1}\frac{3}{4}={\tan }^{-1}x$

${\sin }^{-1}\frac{3}{4}=2{\tan }^{-1}x$

${\tan }^{-1}\left(\frac{\frac{3}{4}}{\sqrt{1-{\left(\frac{3}{4}\right)}^2}}\right)=2{\tan }^{-1}x$

${\tan }^{-1}\left(\frac{3}{\sqrt{7}}\right)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$

$\Rightarrow \frac{3}{\sqrt{7}}=\frac{2x}{1-x^2}$

$3x^2+2\sqrt{7}x-3=0$

$x=\frac{-2\sqrt{7}\pm \sqrt{64}}{6}$

$x=\frac{-2\sqrt{7}\pm 8}{6}$

$x=\frac{-\sqrt{7}\pm 4}{3}$

$x$ cannot be negative because,

$0<{\sin }^{-1}\frac{3}{4}<\frac{\pi }{2}$

$0<\frac{1}{2}{\sin }^{-1}\frac{3}{4}<\frac{\pi }{4}$

$\tan 0<\tan \left(\frac{1}{2}{\sin }^{-1}\frac{3}{4}\right)<\tan \frac{\pi }{4}$

$0<\tan \left(\frac{1}{2}{\sin }^{-1}\frac{3}{4}\right)<1$

So,

$x=\frac{4-\sqrt{7}}{3}$

$\Rightarrow m=7$