Mathematics

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

$2$

##### SOLUTION
$LHS= \dfrac{1}{\sqrt{2k}}\displaystyle\int_0^{\dfrac{\pi}{3}} \dfrac{\tan \theta}{\sqrt{\sec \theta}}d\theta$

$=\dfrac{1}{\sqrt{2k}}\displaystyle \int_0^{\dfrac{\pi}{3}}\dfrac{\sin \theta}{\sqrt{\cos \theta}}d\theta$

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

$=-\dfrac{\sqrt{2}}{\sqrt{k}}\left(\dfrac{1}{\sqrt{2}}-1\right)$

$=\dfrac{\sqrt 2}{\sqrt k}\left(1-\dfrac{1}{\sqrt{2}}\right)$

$\because$ it is given that  $RHS = 1-\dfrac{1}{\sqrt{2}}$

$\therefore$ by comparing we get $k=2$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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