Question

If $y=\log \left(\frac{\sqrt{(x+1)}-1}{\sqrt{(x+1)}+1}\right)+\frac{\sqrt{x}}{\sqrt{(x+1)}}$ the by using substitution $x={\tan }^2\theta ,y$ reduces to

Answer 1

$y=\log \left(\frac{\sqrt{x+1}-1}{\sqrt{x+1}-1}\right)+\frac{\sqrt{x}}{\sqrt{x+1}}$

Given:$x={\tan }^2\theta$

$y=\log \left(\frac{\sqrt{{\tan }^2\theta +1}-1}{\sqrt{{\tan }^2\theta +1}-1}\right)+\frac{\sqrt{{\tan }^2\theta }}{\sqrt{{\tan }^2\theta +1}}$

We know that $1+{\tan }^2\theta ={\sec }^2\theta$

$y=\log \left(\frac{\sqrt{{\sec }^2\theta }-1}{\sqrt{{\sec }^2\theta }-1}\right)+\frac{\sqrt{{\tan }^2\theta }}{\sqrt{{\sec }^2\theta }}$

$y=\log \left(\frac{\sec \theta -1}{\sec \theta +1}\right)+\frac{\tan \theta }{\sec \theta }$

$y=\log \left(\frac{1-\cos \theta }{1+\cos \theta }\right)+\frac{\frac{\sin \theta }{\cos \theta }}{\frac{1}{\cos \theta }}$

We know that $1+\cos \theta =2{\cos }^2\frac{\theta }{2}$ and $1-\cos \theta =2{\sin }^2\frac{\theta }{2}$

$y=\log \frac{{\sin }^2\frac{\theta }{2}}{{\cos }^2\frac{\theta }{2}}+\sin \theta$

$\therefore y=2\log \tan \frac{\theta }{2}+\sin \theta$