What is the Value of \(2\tan\frac{1}{2}A\tan\frac{1}{2}B\) in Triangle ABC?

[Monoxdifly]

Suppose the angles in triangle ABC is A, B, and C. If sin A + sin B = 2 sin C, the value of \(\displaystyle 2tan\frac12Atan\frac12B\) is ...
A. \(\displaystyle \frac83\)
B. \(\displaystyle \sqrt6\)
C. \(\displaystyle \frac73\)
D. \(\displaystyle \frac23\)
E. \(\displaystyle \frac13\sqrt3\)

Since A, B, and C are the angles of triangle ABC, then C = 180° – (A + B)
sin A + sin B = 2 sin C
sin A + sin B = 2 sin(180° – (A + B))
sin A + sin B = 2 sin(A + B)
2 = \(\displaystyle \frac{sinA+sinB}{sin(A+B)}\)

\(\displaystyle 2tan\frac12Atan\frac12B\)
\(\displaystyle \frac{sinA+sinB}{sin(A+B)}×tan\frac12Atan\frac12B\)
What am I supposed to do after this?

 

[skeeter]

$\sin{A} + \sin{B} = 2\sin(A+B)$

$\sin{A} + \sin{B} = 2(\sin{A}\cos{B} + \cos{A}\sin{B}) \implies \cos{B} = \cos{A} = \dfrac{1}{2} \implies A = B = \dfrac{\pi}{3}$

$2\tan^2\left(\dfrac{\pi}{6}\right) = \dfrac{2}{3}$

 

[Monoxdifly]

Thank you.