Question

If $\int\limits^{\frac{\pi}{3}}_0 \frac{\tan \theta}{\sqrt{2 k \sec \theta}} d \theta = 1 - \frac{1}{\sqrt{2}} . (k > 0 )$ , 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^{\pi/3}_{0} \frac{\tan\theta}{\sqrt{\sec\theta}} d \theta = \frac{1}{ \sqrt{2k}} \int^{\pi/3}_{0} \frac{\sin\theta}{\sqrt{\cos\theta}} d \theta $
$ = - \frac{1}{\sqrt{2k}} 2 \sqrt{\cos\theta} ^{\pi/3}_{0} = - \frac{\sqrt{2}}{\sqrt{k}} \left( \frac{1}{\sqrt{2}} - 1\right) $
given it is $1 - \frac{1}{\sqrt{2}} \Rightarrow k = 2 $