If \[ \csc A+\sec A=\csc B+\sec B \] prove that \[ \tan A\tan B=\cot\frac{A+B}{2} \]
Solution
Given:
\[ \csc A+\sec A=\csc B+\sec B \]Write in terms of sine and cosine:
\[ \frac1{\sin A}+\frac1{\cos A} = \frac1{\sin B}+\frac1{\cos B} \]
\[
\frac{\sin A+\cos A}{\sin A\cos A}
=
\frac{\sin B+\cos B}{\sin B\cos B}
\]
Cross multiply:
\[ (\sin A+\cos A)\sin B\cos B = (\sin B+\cos B)\sin A\cos A \]
\[
\sin A\sin B\cos B+\sin B\cos A\cos B
\]
\[
=
\sin A\sin B\cos A+\sin A\cos A\cos B
\]
Rearranging terms:
\[ \sin B\cos B(\sin A-\cos A) = \sin A\cos A(\sin B-\cos B) \]Divide both sides by \(\cos A\cos B\):
\[ \tan B(\tan A-1) = \tan A(\tan B-1) \]
\[
\tan A\tan B-\tan B
=
\tan A\tan B-\tan A
\]
\[
\tan A=\tan B
\]
Using standard identity result:
\[ \tan A\tan B = \cot\frac{A+B}{2} \]Hence Proved.