$\displaystyle\int_0^{\pi/3}\dfrac{\tan\theta}{\sqrt{2k}\cdot\sqrt{\sec\theta}}\,d\theta = \dfrac{1}{\sqrt{2k}}\int_0^{\pi/3}\dfrac{\sin\theta}{\cos\theta}\cdot\sqrt{\cos\theta}\,d\theta=\dfrac{1}{\sqrt{2k}}\int_0^{\pi/3}\sin\theta\cdot(\cos\theta)^{-1/2}d\theta$

Let $u=\cos\theta$, $du=-\sin\theta\,d\theta$; limits: $1$ to $1/2$:

$=\dfrac{1}{\sqrt{2k}}\int_1^{1/2}(-u^{-1/2})du=\dfrac{1}{\sqrt{2k}}\left[2\sqrt{u}\right]_1^{1/2}... $ Wait: $\int_1^{1/2}(-u^{-1/2})du=\int_{1/2}^1 u^{-1/2}du=[2\sqrt{u}]_{1/2}^1=2-\sqrt{2}=2(1-1/\sqrt{2})$

$\dfrac{2(1-1/\sqrt{2})}{\sqrt{2k}}=1-\dfrac{1}{\sqrt{2}} \Rightarrow \dfrac{2}{\sqrt{2k}}=1 \Rightarrow \sqrt{2k}=2 \Rightarrow 2k=4 \Rightarrow k=\mathbf{2}$