Question

If ${\int }_0^{\frac{\pi }{3}}\frac{\tan \theta }{\sqrt{2k\sec \theta }}d\theta =1-\frac{1}{\sqrt{2}},(k>0),$ then the value of k is :

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

Answer 1

The correct option is A $2$

$LHS=\frac{1}{\sqrt{2k}}{\int }_0^{\frac{\pi }{3}}\frac{\tan \theta }{\sqrt{\sec \theta }}d\theta$

$=\frac{1}{\sqrt{2k}}{\int }_0^{\frac{\pi }{3}}\frac{\sin \theta }{\sqrt{\cos \theta }}d\theta$

$=-\frac{1}{\sqrt{2k}}|2\sqrt{\cos \theta }{|}_0^{\frac{\pi }{3}}$

$=-\frac{\sqrt{2}}{\sqrt{k}}(\frac{1}{\sqrt{2}}-1)$

$=\frac{\sqrt{2}}{\sqrt{k}}(1-\frac{1}{\sqrt{2}})$

$\because$ it is given that $RHS=1-\frac{1}{\sqrt{2}}$
$\therefore$ by comparing we get $k=2$