Question

If $A+B={45}^o$ then find $(1+\tan A)(1+\tan B)$+ $(\cot A-1)(\cot B-1)$

Answer 1

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

$\tan (A+B)=\tan {45}^{\circ }$

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

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

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

add 1 on both sides

$1+\tan A+\tan B+\tan A\tan B=2$

$1(1+1\tan A)+\tan B(1+\tan A)=2$

$\frac{(1+\tan A)(1+\tan B)=2}{A+B=45}$

$\cot (A+B)=\cot 45$

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

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

$\cot A+\cot B-\cot A\cot B+1=0$

$\cot A\cot B-\cot A-\cot B-1=0$

adding 2 on both sides

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

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

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