Question

If $A+B={225}^o$, prove that
$\frac{\cot A}{1+\cot A}.\frac{\cot B}{1+\cot B}=\frac{1}{2}$

Answer 1

Changing cot in terms of tan we have to prove that
$\frac{1}{1+\tan A}.\frac{1}{1+\tan B}=\frac{1}{2}$
or $2=1+\tan A+\tan B+\tan A\tan B$
or $\tan A+\tan B=(1-\tan A\tan B)$
or $\frac{\tan A+\tan B}{1-\tan A\tan B}=1or\tan (A+B)=1$
or $\tan {225}^o=1$
But $\tan {225}^o=\tan ({180-}^o+{45}^o)=\tan {45}^o=1$