Question

If $A+B={225}^{\circ }$,
then the value of
$\frac{\cot A}{1+\cot A}.\frac{\cot B}{1+\cot B}$ is

Answer 1

$\frac{\cot A}{1+\cot A}.\frac{\cot B}{1+\cot B}=\frac{1}{(1+\tan A)(1+\tan B)}$
$=\frac{1}{\tan A+\tan B+1+\tan A\tan B}$

Given $A+B={225}^{\circ }$

$\Rightarrow \tan (A+B)=\tan {225}^{\circ }$

$\Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B}=\tan (180+45{)}^{\circ }$
$\Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B}=1$

$\Rightarrow \tan A+\tan B=1-\tan A\tan B$

$\Rightarrow \tan A+\tan B+1+\tan A\tan B=1+1$

$\Rightarrow \frac{1}{1+\tan A\tan B+1+\tan A\tan B}=\frac{1}{2}$

Hence the correct answer is Option D.