Question:
The number of values of $\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ satisfying $\frac{1-\cos 2 \theta}{1+\cos 2 \theta}=3$ is ____________________
Solution:
for $\theta \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$
$\frac{1-\cos 2 \theta}{1+\cos 2 \theta}=3$
i.e $\frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta}=3 \quad$ (using identity :- $1-\cos ^{2} \theta=2 \sin ^{2} \theta, 1+\cos 2 \theta=2 \cos ^{2} \theta$ )
i.e $\tan ^{2} \theta=3$
i.e $\tan \theta=\pm \sqrt{3}$
i.e $\theta=\frac{\pi}{3}, \pi-\frac{\pi}{3}, \pi+\frac{\pi}{3}, 2 \pi-\frac{\pi}{3}, 2 \pi+\frac{\pi}{3}$
out of which $\frac{\pi}{3}, \frac{2 \pi}{3}$ lie in $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$
Hence, number of value $\theta$ is 2 .