Question

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

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

Answer 1

The correct option is A $2$

$\underset{0}{\overset{\pi /3}{\int }}\frac{\tan \theta }{\sqrt{2k\sec \theta }}d\theta =1-\frac{1}{\sqrt{2}}$

$\Rightarrow \underset{0}{\overset{\pi /3}{\int }}\frac{\tan \theta }{\sqrt{\sec \theta }}d\theta =\sqrt{2k}(1-\frac{1}{\sqrt{2}})\cdots \left(1\right)$

Let $I=\underset{0}{\overset{\pi /3}{\int }}\frac{\tan \theta }{\sqrt{\sec \theta }}d\theta$
Put $\sec \theta =t^2$
$\Rightarrow \sec \theta \tan \theta d\theta =2tdt$
$\Rightarrow \tan \theta d\theta =\frac{2}{t}dt$
When $\theta =0,t=1$
When $\theta =\frac{\pi }{3},t=\sqrt{2}$
$\therefore I=\underset{1}{\overset{\sqrt{2}}{\int }}\frac{2}{t^2}dt$
$=2{(-\frac{1}{t})}_1^{\sqrt{2}}$
$=2(-\frac{1}{\sqrt{2}}+1)$
$\Rightarrow I=2(1-\frac{1}{\sqrt{2}})$

From Equation $\left(1\right)$,
$2(1-\frac{1}{\sqrt{2}})=\sqrt{2}\sqrt{k}(1-\frac{1}{\sqrt{2}})$
$\Rightarrow \sqrt{k}=\sqrt{2}$
$\Rightarrow k=2$