Question

If cosec A + sec A = cosec B + sec B, prove that tan A tan B = $\cot \frac{A+B}{2}$.

Answer 1

Given:
$$\begin{gathered} \frac{1}{\sin A}+\frac{1}{\cos A}=\frac{1}{\sin B}+\frac{1}{\cos B} \\ \Rightarrow \frac{1}{\sin A}-\frac{1}{\sin B}=\frac{1}{\cos B}-\frac{1}{\cos A} \\ \Rightarrow \frac{\sin B-\sin A}{\sin A\sin B}=\frac{\cos A-\cos B}{\cos A\cos B} \\ \Rightarrow \frac{\sin B-\sin A}{\cos A-\cos B}=\frac{\sin A\sin B}{\cos A\cos B} \\ \Rightarrow \frac{2\sin \left({\displaystyle \frac{B-A}{2}}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)}{-2\sin \left({\displaystyle \frac{A-B}{2}}\right)\sin \left({\displaystyle \frac{A+B}{2}}\right)}=\frac{\sin A\sin B}{\cos A\cos B} \\ \Rightarrow \frac{-\sin \left({\displaystyle \frac{A-B}{2}}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)}{-\sin \left({\displaystyle \frac{A-B}{2}}\right)\sin \left({\displaystyle \frac{A+B}{2}}\right)}=\frac{\sin A\sin B}{\cos A\cos B} \\ \Rightarrow \frac{\cos \left({\displaystyle \frac{A+B}{2}}\right)}{\sin \left({\displaystyle \frac{A+B}{2}}\right)}=\frac{\sin A\sin B}{\cos A\cos B} \\ \Rightarrow \cot \left(\frac{A+B}{2}\right)=\tan A\tan B \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$