Question

16. If A and B are acute angles such that tan A=1/3
tanB=1/2 and
tan (A+B)=tanA+tanB/1-tanATanB
show that A+B=45°.
​

Answer 1

Answer:

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

Also,

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

1−tanAtanB

tanA+tanB

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

1−

3

1

×

2

1

3

1

+

2

1

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

6

6−1

6

2+3

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

6

5

6

5

\implies tan(A+B) = 1⟹tan(A+B)=1

\implies (A+B) = tan^{-1} 1⟹(A+B)=tan

−1

1

\implies A + B = 45^{\circ}⟹A+B=45

∘

( Since, tan^{-1} 1 = 45^{\circ}tan

−1

1=45

∘

)