If $\tan \frac{\theta}{2}=\operatorname{cosec} \theta-\sin \theta$, then $\tan ^2 \frac{\theta}{2}=$

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Aicharya · Accepted Answer

Given, $\tan \frac{\theta}{2}=\operatorname{cosec} \theta-\sin \theta$ $\begin{aligned} & =\frac{1}{\sin \theta}-\sin \theta=\frac{1-\sin ^2 \theta}{\sin \theta} \ & =\frac{\cos ^2 \theta}{\sin \theta}=\cot \theta \cos \theta=\frac{\left(1-\tan ^2 \frac{\theta}{2}\right)^2}{2 \tan \frac{\theta}{2}\left(1+\tan ^2 \frac{\theta}{2}\right)} \end{aligned}$ $\begin{aligned} & \Rightarrow \quad 2 \tan ^2 \frac{\theta}{2}\left(1+\tan ^2 \frac{\theta}{2}\right)=1+\tan ^4 \frac{\theta}{2}-2 \tan ^2 \frac{\theta}{2} \ & \Rightarrow \quad 2 \tan ^2 \frac{\theta}{2}+2 \tan ^4 \frac{\theta}{2}=1+\tan ^4 \frac{\theta}{2}-2 \tan ^2 \frac{\theta}{2} \ & \Rightarrow \quad \tan ^4 \frac{\theta}{2}+4 \tan ^2 \frac{\theta}{2}-1=0 \ & \therefore \quad \tan ^2 \frac{\theta}{2}=\frac{-4 \pm \sqrt{(4)^2-(-1) 4}}{2 \times 1}=\frac{-4 \pm \sqrt{16+4}}{2} \ & \quad=\frac{-4 \pm \sqrt{20}}{2}=-2 \pm \sqrt{5} \quad\left[\because \tan ^2 \frac{\theta}{2} > 0\right] \end{aligned}$

If \(\tan \frac{\theta}{2}=\operatorname{cosec} \theta-\sin \theta\), then \(\tan ^2 \frac{\theta}{2}=\)

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