Question

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

Answer 1

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