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 $$2tan\frac12Atan\frac12B$$ is ...
A. $$\frac83$$
B. $$\sqrt6$$
C. $$\frac73$$
D. $$\frac23$$
E. $$\frac13\sqrt3$$

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 = $$\frac{sinA+sinB}{sin(A+B)}$$

$$2tan\frac12Atan\frac12B$$
$$\frac{sinA+sinB}{sin(A+B)}×tan\frac12Atan\frac12B$$
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.