Question

If $A+B={45}^0,$ then prove that
(a) $(1+\tan A)(1+\tan B)=2$
(b) $(\cot A-1)(\cot B-1)=2$
(c) $A-B=\frac{3\pi }{4},$ then show that $(1-\tan A)(1+\tan B)=2$

Answer 1

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

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

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

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

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

$\therefore (1+\tan A)(1+\tan B)=2\to$ $\left(a\right)$

$\cot (A+B)=1$

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

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

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

$\Rightarrow (\cot A-1)(\cot B-1)=2\to$ $\left(b\right)$

$A-B=\frac{3\pi }{4}$

$\Rightarrow \tan (A-B)=-1$

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

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

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

$\Rightarrow 2=(1-\tan A)(1+\tan B)\to$ $\left(c\right)$