Question

If $\underset{0}{\overset{\frac{\pi }{3}}{\int }}\frac{\tan \theta }{\sqrt{2k\sec \theta }}d\theta =1-\frac{1}{\sqrt{2}}.(k>0)$ , then the value of $k$ is :

• A. 2
• B. $\frac{1}{2}$
• C. 4
• D. 1

Answer 1

Answer: (A)

$\frac{1}{\sqrt{2k}}{\int }_0^{\pi /3}\frac{\tan \theta }{\sqrt{\sec \theta }}d\theta =\frac{1}{\sqrt{2k}}{\int }_0^{\pi /3}\frac{\sin \theta }{\sqrt{\cos \theta }}d\theta$
$=-\frac{1}{\sqrt{2k}}2{\sqrt{\cos \theta }}_0^{\pi /3}=-\frac{\sqrt{2}}{\sqrt{k}}\left(\frac{1}{\sqrt{2}}-1\right)$
given it is $1-\frac{1}{\sqrt{2}}\Rightarrow k=2$