Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

$\frac{{\sin }^{-1}x}{r}=a,\frac{{\cos }^{-1}x}{r}=b,\frac{{\tan }^{-1}y}{r}=c$

So, $a+b=\frac{\pi }{2r}$

$\cos \left(\frac{\pi c}{a+b}\right)=\cos \left(\frac{\pi {\tan }^{-1}y}{\frac{\pi }{2r}r}\right)$

$=\cos \left(2{\tan }^{-1}y\right),$ let ${\tan }^{-1}y=\theta$

$=\cos \left(2\theta \right)$

$=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\frac{1-y^2}{1+y^2}$