Trig2.13

If A+B=45, prove that $(1+\tan A)(1+\tan B)=2\,$ .Hence show that $\tan \frac{45}{2} =\sqrt{2}-1\,$

Given A+B=45

$\tan (A+B)=\tan 45\,$

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

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

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

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

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

If $A=B=\frac{45}{2}\,$ then $A+B=45\,$ .Hence $(1+\tan A)(1+\tan B)=2\,$

$[1+\tan \frac{45}{2}][1+\tan \frac{45}{2}]=2\,$ implies $[1+\tan \frac{45}{2}]=2\,$

$1+\tan \frac{45}{2}=\pm \sqrt{2},\tan \frac{45}{2}=\pm \sqrt{2}-1\,$

Therefore $\tan \frac{45}{2}=\pm \sqrt{2}-1\,$