Question

If $\frac{\pi }{2}<\theta <\pi$, then write the value of $\sqrt{\frac{1-cos2\theta }{1+cos2\theta }}$

Answer 1

Since $\frac{\pi }{2}<\theta <\pi$

$\Rightarrow \theta$ lies in the $2^{nd}$ quadrant

Now,

$\sqrt{\frac{1-cos2\theta }{1+cos2\theta }}\Rightarrow \sqrt{\frac{2si{n}^2\theta }{2co{s}^2\theta }}\Rightarrow \frac{-sin\theta }{-cos\theta }(\because \theta \text{lies in the}{2}^{nd}\text{quadrant})\Rightarrow -tan\theta$