Question

If $A+B=\frac{\pi }{4}$, then $(\cot A-1)(\cot B-1)$ is equal to

• A. $18$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is B $2$

We have, $A+B=\frac{\pi }{4}$

$\tan (A+B)=\tan \frac{\pi }{4}$

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

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

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

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

$\cot B+\cot A+1=\cot A\cot B$

$\cot A\cot B-\cot A-\cot B=1$

$\cot A\cot B-\cot A-\cot B+1=2$

$\cot A(\cot B-1)-(\cot B-1)=2$

$(\cot A-1)(\cot B-1)=2$