If \(\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}, 0 \leq \alpha \leq \frac{\pi}{2}\), then the value of \(\cos 2 \theta\) is

\(\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha} \)
\( \frac{\sin \theta}{\cos \theta}=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha} \)
\( \therefore \sin \alpha \sin \theta+\cos \alpha \sin \theta=\sin \alpha \cos \theta-\cos \alpha \cos \theta \)
\( \therefore \cos \alpha \cos \theta+\sin \alpha \sin \theta=\sin \alpha \cos \theta-\cos \alpha \sin \theta \)
\( \therefore \cos (\alpha-\theta)=\sin (\alpha-\theta) \)
\( \therefore \alpha-\theta=\frac{\pi}{4} \)
\( \therefore \theta=\alpha-\frac{\pi}{4} \)
\( \therefore 2 \theta=2 \alpha-\frac{\pi}{2} \)
\( \therefore \cos 2 \theta=\cos \left(2 \alpha-\frac{\pi}{2}\right) \)
\( =\cos \left[\because 0 \leq \alpha \leq \frac{\pi}{2}\right] \)
\( \left.=\cos \left(\frac{\pi}{2}-2 \alpha\right)\right] \ldots[\because \cos (-\theta)=\cos \theta] \)
\( \therefore \cos 2 \theta=\sin 2 \alpha\)