Question

If $A+B={45}^{\circ }$, prove that $(1+\tan A)(1+\tan B)=2$ and hence deduce that $\tan 22{\frac{1}{2}}^{\circ }=\sqrt{2}-1$

Answer 1

$\to A+B={45}^0$

$\to A={45}^0-B$

$\to \tan A=\tan ({45}^0-B)=\frac{1-\tan B}{1+\tan B}$

$\to \tan A+1=\frac{1-\tan B}{1+\tan B}+1=\frac{2}{1+\tan B}$

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

Now, for $A=B={22}_2^1$

$\to {(1+\tan A)}^2=2$

$\to \tan A=\tan \left({22}_2^1\right)=\sqrt{2}-1$