Question

If
\(\displaystyle\frac{\sin^{-1}x}{a}=\frac{\cos^{-1}x}{b}=\frac{\tan^{-1}y}{c}\)
;
\(0\lt x\lt1\)
, then the value of
\(\cos\left(\frac{\pi c}{a+b}\right)\)
is :

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

Hint:

Add the first two ratios using \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\), then express \(\frac{\pi c}{a+b}\) in terms of \(\tan^{-1}y\).

Answer 1

The Correct Option is D

Let the common ratio be \(k\): \(\sin^{-1}x=ka\), \(\cos^{-1}x=kb\), \(\tan^{-1}y=kc\).
Adding: \(k(a+b)=\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\), so \(k=\frac{\pi}{2(a+b)}\).
Then \(\tan^{-1}y=kc=\frac{\pi c}{2(a+b)}\), i.e. \(\frac{\pi c}{a+b}=2\tan^{-1}y\).
\(\cos\left(2\tan^{-1}y\right)=\frac{1-y^2}{1+y^2}\).
Answer: (4) \(\frac{1-y^2}{1+y^2}\)