Question

Prove: $\frac{(cotA+\tan B)}{(cotB+\tan A)}=cotA\tan B$

Answer 1

Solve the L.H.S part:

$$\begin{gathered} \frac{(cotA+\tan B)}{(cotB+\tan A)}=\frac{({\displaystyle \frac{1}{\tan A}}+\tan B)}{({\displaystyle \frac{1}{\tan B}}+\tan A)} \\ \Rightarrow =\frac{\left({\displaystyle \frac{1+\tan B\tan A}{\tan A}}\right)}{({\displaystyle \frac{1+\tan A\tan B}{\tan B}})} \\ \Rightarrow =\frac{1+\tan B\tan A}{\tan A}\times \frac{\tan B}{1+\tan A\tan B} \\ \Rightarrow =\frac{1}{\tan A}\times \frac{\tan B}{1} \\ \Rightarrow =cotA\tan B \end{gathered}$$

Since, L.H.S = R.H.S.

Hence, proved.