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 :


ANSWER

$$2$$


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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