Question

If A and B are acute angles such that $tan A=\frac{1}{2}$, $tan B=\frac{1}{3}$ and $tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}$, find A + B.

Answer 1

SOLUTION :

Given :  

If A & B are acute angles , tan A = 1/2 , tan B =1/3 and tan(A+B) = tanA + tanB/1− tanA tanB

tan(A+B) = tanA + tanB /1− tanA tanB

tan(A+B) =½ +⅓ / (1− ½  × ⅓)

[ Given : tan A = 1/2 , tan B =⅓]

tan(A+B) = [(3+2)/6] / (1 - ⅙)

[ By taking L.C.M]

tan(A+B) = ⅚ / (6-1)/6

[ By taking L.C.M]

tan(A+B) = ⅚ / ⅚

tan(A+B) = ⅚ × 6/5  = 1

tan(A+B) = 1

tan(A+B) = tan 45°      [tan 45° = 1]

(A+B) = 45°

Hence, the value of (A+B) = 45°.  

HOPE THIS ANSWER WILL HELP YOU…