Question
Prove the following : $\frac{\tan \left(\frac{\theta}{2}\right)+\cot \left(\frac{\theta}{2}\right)}{\cot \left(\frac{\theta}{2}\right)-\tan \left(\frac{\theta}{2}\right)}=\sec \theta$

Answer

$\begin{aligned}
& \text { L.H.S. }=\frac{\tan \frac{\theta}{2}+\cot \frac{\theta}{2}}{\cot \frac{\theta}{2}-\tan \frac{\theta}{2}} \\
& =\frac{\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}+\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}}{\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}-\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}} \\
& =\frac{\sin ^2 \frac{\theta}{2}+\cos ^2 \frac{\theta}{2}}{\cos ^2 \frac{\theta}{2}-\sin ^2 \frac{\theta}{2}} \\
& =\frac{1}{\cos \theta} \\
& =\sec \theta=\text { R.H.S. } \\
\end{aligned}$

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