Question

If $A+B={45}^o$, show that,$(1+tanA)(1+tanB)=2$; hence find the value of $\tan {\left(22\frac{1}{2}\right)}^o$

Answer 1

Given $A+B=45$

{Take tan on both the sides }

$tan(A+B)=tan45$

$\frac{tanA+tanB}{1-tanAtanB}=1$

$tanA+tanB=1-tanA.tanB$

$tanA+tanB+tanA.tanB=1$

adding "1" on both sides

$1+tanA+tanB+tanA.tanB=1+1$

$(1+tanA)+tanB(1+tanA)=2$

$(1+tanA)(1+tanB)=2$

putting $A=B=22.5$

$(1+tan22.5)(1+tan22.5)=2$

or ${(1+tan22.5)}^2=2$

or $1+tan22.5=\sqrt{2}$

as it is positive

or $tan22.5=\sqrt{2}-1$