Mathematics - Inverse Trigonometric Function Question with Solution | TestHub

Let $\frac{1}{4}{\sin }^{-1}\frac{\sqrt{63}}{8}=\theta$

$\sin 4\theta =\frac{\sqrt{63}}{8}$

$\cos 4\theta =\frac{1}{8}$

$2{\cos }^{2}2\theta -1=\frac{1}{8}$

${\cos }^{2}2\theta =\frac{9}{16}$

$\cos 2\theta =\frac{3}{4}$

$2{\cos }^2\theta -1=\frac{3}{4}$

${\cos }^2\theta =\frac{7}{8}$

$\cos \theta =\frac{\sqrt{7}}{2\sqrt{2}}$

$\tan \theta =\frac{1}{\sqrt{7}}$