Question

PLEASE SOLVE THIS QUESTION....

Answer 1

hope this will help you.. :-)

Answer 2

$\huge\bold {\red {solution}}$
$\frac{ \tan A + \cot B}{ \cot A + \tan B } \\ \\ = \frac{\frac{1}{\cot B} + \cot A}{\cot A + \frac{1}{\cot B}} \\ \\ = \frac{\frac{\cot A .\cot B + 1}{\cot B}}{\frac{\cot A . \cot B + 1}{\cot A}} \\ \\ = \frac{\cot B}{\cot A} \\ \\ = \cot B . \tan A$