If \(\int_0^{\frac{\pi}{3}} \frac{\tan \theta}{\sqrt{2 \mathrm{ksec} \theta}} \mathrm{d} \theta=1-\frac{1}{\sqrt{2}},(k\gt0)\), then the value of \(k\) is

\( \text { Let } \mathrm{I}=\int_0^{\frac{\pi}{3}} \frac{\tan \theta}{\sqrt{2 \mathrm{k} \sec \theta}} \mathrm{d} \theta \)
\( =\frac{1}{\sqrt{2 \mathrm{k}}} \int_0^{\frac{\pi}{3}} \frac{\sin \theta}{\cos \theta} \times \sqrt{\cos \theta} \mathrm{d} \theta \)
\( =\frac{1}{\sqrt{2 \mathrm{k}}} \int_0^{\frac{\pi}{3}} \frac{\sin \theta}{\sqrt{\cos \theta}} \mathrm{~d} \theta\)
\(\therefore \text { Let } \cos \theta=\mathrm{t} \)
\( \sin \theta \mathrm{d} \theta=-\mathrm{dt}\)
\( \therefore \text { when } \theta=\frac{\pi}{3}, \mathrm{t}=\frac{1}{2} \)
\( \text { when } \theta=0, \mathrm{t}=1 \)
\( \therefore I=\frac{-1}{\sqrt{2 \mathrm{k}}} \int_1^{\frac{1}{2}} \frac{1}{\sqrt{\mathrm{t}}} \mathrm{dt} \)
\( =\frac{-1}{\sqrt{2 \mathrm{k}}} \frac{\left.\mathrm{t}^{\frac{1}{2}}\right]_1^{\frac{1}{2}}}{\frac{1}{2}}\)
\(=\frac{-\sqrt{2}}{\sqrt{\mathrm{k}}}\left(\frac{1}{\sqrt{2}}-1\right)=\frac{\sqrt{2}}{\sqrt{\mathrm{k}}}\left(1-\frac{1}{\sqrt{2}}\right)\)
Given that \(I=\left(1-\frac{1}{\sqrt{2}}\right)\)
\(\Rightarrow \frac{\sqrt{2}}{\sqrt{\mathrm{k}}}=1 \Rightarrow \mathrm{k}=2\)