Question

If $A+B=\frac{\pi }{4}$ then prove that $(\cot A-1)(\cot B-1)=2$.

Answer 1

We are given $A+B=\frac{\pi }{4}$

now, $\cot (A+B)=\cot \left(\frac{\pi }{4}\right)$

$\therefore \frac{\cot A\cot B-1}{\cot A+\cot B}=1$

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

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

Now, $LHS=(\cot A-1)(\cot B-1)$

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

$=1+1$

$=2$

$=RHS$

Hence, Prove that $(\cot A-I)(\cot B-I)=2$