Question

If $A+B={45}^o$, prove that $(\cot A-1)(\cot B-1)=2$

Answer 1

Given $A+B={45}^{\circ }$

$⟹\cot (A+B)=\cot {45}^{\circ }$

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

$⟹\cot A\cot B-1=\cot A+\cot B$

$⟹\cot A\cot B-\cot A-\cot B=1$

$⟹\cot A\cot B-\cot A-\cot B+1=2$

$⟹\cot A(\cot B-1)-(\cot B-1)=2$

$⟹(\cot A-1)(\cot B-1)=2$

Hence proved.