Question

If $\frac{{\sin }^{-1}x}{a}=\frac{{\cos }^{-1}x}{b}=\frac{{\tan }^{-1}y}{c};0<x<1,$ then the value of $\cos \left(\frac{\pi c}{a+b}\right)$ is

• A. $\frac{1-y^2}{y\sqrt{y}}$
• B. $1-y^2$
• C. $\frac{1-y^2}{1+y^2}$
• D. $\frac{1-y^2}{2y}$

Answer 1

Answer: (C)

$\frac{{\sin }^{-1}x}{r}=a,\frac{{\cos }^{-1}x}{r}=b,\frac{{\tan }^{-1}y}{r}=c$
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 ${\tan }^{-1}y=\theta$
$=\cos (2\theta )$
$=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\frac{1-y^2}{1+y^2}$