Question

The number of values of $\theta \in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$ satisfying $\frac{1-\cos 2\theta }{1+\cos 2\theta }=3$ is ______________.

Answer 1

$$\begin{gathered} \mathrm{for}\theta \in \left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right) \\ \frac{1-\cos 2\theta }{1+\cos 2\theta }=3 \end{gathered}$$

i.e $\frac{2{\sin }^2\theta }{2{\cos }^2\theta }=3$ (using identity :- 1 − cos²$\theta$ = 2 sin²$\theta$, 1 + cos 2$\theta$ = 2cos²$\theta$)
i.e tan²$\theta$ = 3
i.e tan $\theta$ = $\pm \sqrt{3}$
i.e $\theta =\frac{\mathrm{\pi }}{3},\mathrm{\pi }-\frac{\mathrm{\pi }}{3},\mathrm{\pi }+\frac{\mathrm{\pi }}{3},2\mathrm{\pi }-\frac{\mathrm{\pi }}{3},2\mathrm{\pi }+\frac{\mathrm{\pi }}{3}$
out of which $\frac{\mathrm{\pi }}{3},\frac{2\mathrm{\pi }}{3}$ lie in $\left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$
Hence, number of value $\theta$ is 2.