Question

if A and B are acute angles such that tan A= 1/3 , tan B = 1/2 and tan( A +B) = tan A+ tanB / 1- tanA tanB, show that A+ B = 45 degree

Answer 1

using the formula
Tan(A+B)=tanA+TanB/1-tanA.tanB
putting the value of tanA &tanB
tan(A+B)=1/3+1/2upon1-1/2.1/3
tan(A+B)=1
tan(A+B)=tan45 degree
A+B=45 degree
hence proved

Answer 2

Answer:

Here, tan A = 1/3, tan B = 1/2

Also,

$tan( A +B) = \frac{tan A+ tanB}{1- tanA tanB}$

$\implies tan(A+B)=\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\times \frac{1}{2}}$

$\implies tan(A+B) =\frac{\frac{2+3}{6}}{\frac{6-1}{6}}$

$\implies tan(A+B) = \frac{\frac{5}{6}}{\frac{5}{6}}$

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

$\implies (A+B) = tan^{-1} 1$

$\implies A + B = 45^{\circ}$     ( Since, $tan^{-1} 1 = 45^{\circ}$  )

Hence, proved.