Question

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

• A. $\log {\tan }^2\frac{\theta }{4}+\sin \theta$
• B. $\log {\tan }^2\frac{\theta }{2}+\sin \theta$
• C. $\log {\tan }^2\theta +\sin \theta$
• D. $\log {\tan }^2\frac{\theta }{2}+\sin \frac{\theta }{2}$

Answer 1

The correct option is B $\log {\tan }^2\frac{\theta }{2}+\sin \theta$

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

Using substitution, $x={\tan }^2\theta$, we get

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

Hence, option B.