If (\int_0^{\frac{\pi}{3}} \tan\theta \sqrt{2k\sec\theta} d\theta = \frac{1}{\sqrt{2}}, (k>0)), then the value of k is :

LHS = \frac{1}{\sqrt{2k}} \int_0^{\frac{\pi}{3}} \tan\theta \sqrt{\sec\theta} d\theta = \frac{1}{\sqrt{2k}} \int_0^{\frac{\pi}{3}} \frac{\sin\theta}{\cos\theta} \frac{1}{\sqrt{\cos\theta}} d\theta = \frac{1}{\sqrt{2k}} \int_0^{\frac{\pi}{3}} \frac{\sin\theta}{\cos^{3/2}\theta} d\theta
Let u = \cos\theta \implies du = -\sin\theta d\theta
When (\theta = 0, u = 1)
When (\theta = \frac{\pi}{3}, u = \frac{1}{2})
Then LHS = \frac{1}{\sqrt{2k}} \int_1^{\frac{1}{2}} -u^{-3/2} du = \frac{1}{\sqrt{2k}} \int_{\frac{1}{2}}^1 u^{-3/2} du = \frac{1}{\sqrt{2k}} [ \frac{u^{-1/2}}{-1/2} ]{\frac{1}{2}}^1 = \frac{1}{\sqrt{2k}} [-2u^{-1/2}]{\frac{1}{2}}^1 = \frac{1}{\sqrt{2k}} [-2(1 - \sqrt{2})] = \frac{2(\sqrt{2} - 1)}{\sqrt{2k}}
It is given that RHS = \frac{1}{\sqrt{2}}
Therefore, \frac{2(\sqrt{2} - 1)}{\sqrt{2k}} = \frac{1}{\sqrt{2}}
\implies \sqrt{2k} = 2\sqrt{2}(\sqrt{2} - 1)
\implies 2k = 8 - 4\sqrt{2}
\implies k = 4 - 2\sqrt{2}
However, this solution is incorrect. Let's try a different approach.
\frac{1}{\sqrt{2k}} \int_0^{\pi/3} \frac{\sin\theta}{\cos^{3/2}\theta} d\theta = \frac{1}{\sqrt{2}}
Let u = \cos\theta, du = -\sin\theta d\theta
\frac{1}{\sqrt{2k}} \int_1^{1/2} -u^{-3/2}du = \frac{1}{\sqrt{2k}} [2u^{-1/2}]{1/2}^1 = \frac{1}{\sqrt{2k}} (2 - 2\sqrt{2}) = \frac{2(1-\sqrt{2})}{\sqrt{2k}} = \frac{1}{\sqrt{2}}
\sqrt{2k} = 2\sqrt{2}(1-\sqrt{2}) = 2\sqrt{2} - 4
2k = 8 - 8\sqrt{2}
k = 4 - 4\sqrt{2}
This is still incorrect.
Let's re-examine the integral.
\int_0^{\pi/3} \tan\theta \sqrt{2k\sec\theta} d\theta = \int_0^{\pi/3} \frac{\sin\theta}{\cos\theta} \sqrt{\frac{2k}{\cos\theta}} d\theta = \sqrt{2k} \int_0^{\pi/3} \frac{\sin\theta}{\cos^{3/2}\theta} d\theta = \frac{1}{\sqrt{2}}
Let u = \cos\theta, du = -\sin\theta d\theta
\sqrt{2k} \int_1^{1/2} -u^{-3/2} du = \sqrt{2k} [2u^{-1/2}]{1/2}^1 = \sqrt{2k} (2 - 2\sqrt{2}) = \frac{1}{\sqrt{2}}
\sqrt{2k} (2 - 2\sqrt{2}) = \frac{1}{\sqrt{2}}
\sqrt{2k} = \frac{1}{\sqrt{2}(2 - 2\sqrt{2})} = \frac{1}{2\sqrt{2}(1 - \sqrt{2})} = \frac{\sqrt{2}}{2(1-\sqrt{2})} = \frac{\sqrt{2}}{2(1-\sqrt{2})} \times \frac{1+\sqrt{2}}{1+\sqrt{2}} = \frac{\sqrt{2}(1+\sqrt{2})}{2(-1)} = -\frac{\sqrt{2}+2}{2}
This gives a negative k, which is not possible. There must be a mistake in the problem statement or solution.