Question

If $\pi <\mathrm{\theta }<\frac{3\pi }{2}$, then write the value of $\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \sqrt{\frac{1-\cos 2\mathrm{\theta }}{1+\cos 2\mathrm{\theta }}}=\sqrt{\frac{2{\sin }^2\mathrm{\theta }}{2{\cos }^2\mathrm{\theta }}} \\ =\frac{\left|\mathrm{sin\theta }\right|}{\left|\mathrm{cos\theta }\right|} \\ =\frac{\left|\mathrm{sin\theta }\right|}{\left|\mathrm{cos\theta }\right|} \\ =\frac{-\mathrm{sin\theta }}{-\mathrm{cos\theta }}\left(\because \mathrm{\pi }<\mathrm{\theta }<\frac{3\mathrm{\pi }}{2}\right) \\ \therefore \sqrt{\frac{1-\cos 2\mathrm{\theta }}{1+\cos 2\mathrm{\theta }}}=\mathrm{tan\theta } \end{gathered}$$