Question

P.T. $\frac{cotA+tanB}{cotB+tanA}=cotAtanB$

Answer 1

L.H.S$=\frac{\cot A+\tan B}{\cot B+\tan A}$

$=\frac{\frac{1}{\tan A}+\tan B}{\frac{1}{\tan B}+\tan A}$

$=\frac{\frac{1+\tan A\tan B}{\tan A}}{\frac{1+\tan A\tan B}{\tan B}}$

$=\frac{\tan B}{\tan A}=\tan B\cot A=$R.H.S

Hence proved.