Question

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

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

Answer 1

The correct option is B

$\frac{1-y^2}{1+y^2}$

Explanation for the correct option.

Step 1: Finding the value of $(a+b)$

Given that, $\frac{\left({\sin }^{-1}x\right)}{a}=\frac{\left({\cos }^{-1}x\right)}{b}=\frac{\left({\tan }^{-1}y\right)}{c};0<x<1$

Let, $\frac{\left({\sin }^{-1}x\right)}{a}=\frac{\left({\cos }^{-1}x\right)}{b}=\frac{\left({\tan }^{-1}y\right)}{c}=k$

$\begin{array}{c}\therefore a+b=\frac{{\sin }^{-1}x}{k}+\frac{{\cos }^{-1}x}{k}\\ =\frac{1}{k}({\sin }^{-1}x+{\cos }^{-1}x)[\because 0<x<1]\\ \\ =\frac{1}{k}\times \frac{\pi }{2}\\ =\frac{\pi }{2k}\end{array}$

Step 2: Finding the value of $\cos \left(\frac{\pi c}{a+b}\right)$

Substitute the value of $\left(a+b\right)$

$\begin{array}{c}\therefore \cos \left(\frac{\pi c}{a+b}\right)=cos\left(\frac{\pi {\tan }^{-1}y}{\frac{\pi }{2k}\times k}\right)\left[\because c=\frac{{\tan }^{-1}x}{k}\right]\\ =cos\left(2{\tan }^{-1}y\right)\left[\because {\text{Let, tan}}^{-1}y=\theta \Rightarrow \tan \theta =y\right]\\ =\cos 2\theta \end{array}$

Put $\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{\theta }\mathbf{=}\frac{\mathbf{1}\mathbf{-}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\mathbf{\theta }}{\mathbf{1}\mathbf{+}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\mathbf{\theta }}$

$\begin{array}{c}\therefore \cos \left(\frac{\pi c}{a+b}\right)=\cos 2\theta \\ =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\\ =\frac{1-y^2}{1+y^2}\end{array}$

Hence, the correct option is (B)