If intlimits_0^{frac{pi }{3}} {frac{{tan thet | vedclass

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(A) Given the integral $I = \int\limits_0^{\frac{\pi }{3}} \frac{	an 	heta}{\sqrt{2k \sec 	heta}} d	heta = 1 - \frac{1}{\sqrt{2}}$.We can rewrite

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$2$

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(A) Given the integral $I = \int\limits_0^{\frac{\pi }{3}} \frac{	an 	heta}{\sqrt{2k \sec 	heta}} d	heta = 1 - \frac{1}{\sqrt{2}}$.We can rewrite the integrand as:$I = \frac{1}{\sqrt{2k}} \int\limits_0^{\frac{\pi }{3}} \frac{\sin 	heta / \cos 	heta}{\sqrt{1 / \cos 	heta}} d	heta = \frac{1}{\sqrt{2k}} \int\limits_0^{\frac{\pi }{3}} \frac{\sin 	heta}{\sqrt{\cos 	heta}} d	heta$.Let $u = \cos 	heta$,then $du = -\sin 	heta d	heta$. When $	heta = 0, u = 1$. When $	heta = \frac{\pi}{3}, u = \frac{1}{2}$.$I = \frac{1}{\sqrt{2k}} \int_{1}^{1/2} -u^{-1/2} du = \frac{1}{\sqrt{2k}} \int_{1/2}^{1} u^{-1/2} du$.$I = \frac{1}{\sqrt{2k}} [2\sqrt{u}]_{1/2}^{1} = \frac{2}{\sqrt{2k}} (1 - \frac{1}{\sqrt{2}}) = \sqrt{\frac{2}{k}} (1 - \frac{1}{\sqrt{2}})$.Equating this to the given value $1 - \frac{1}{\sqrt{2}}$:$\sqrt{\frac{2}{k}} (1 - \frac{1}{\sqrt{2}}) = 1 - \frac{1}{\sqrt{2}}$.Thus,$\sqrt{\frac{2}{k}} = 1$,which implies $\frac{2}{k} = 1$,so $k = 2$.

If $\int\limits_0^{\frac{\pi }{3}} {\frac{{\tan \theta }}{{\sqrt {2k \sec \theta } }}} \,d\theta = 1 - \frac{1}{\sqrt{2}}, (k > 0),$ then the value of $k$ is

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