Problem 2

a

b

, and

c

are the sides and

\alpha

\beta

, and

\gamma

are the respective angles of a triangle for which

a+b=\tan\frac{\gamma}{2}\left(a\tan\alpha+b\tan\beta\right)

, prove that the triangle is isosceles.

Step 1 of 6: Clear the tangents from the hypothesis

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In plain words

Turning a mixed tangent identity into one polynomial equation in sines and cosines makes the angle-addition formulas usable in the next step.

(a+b)\cos\alpha\cos\beta\cos\frac{\gamma}{2} = a\sin\alpha\cos\beta\sin\frac{\gamma}{2} + b\cos\alpha\sin\beta\sin\frac{\gamma}{2}

Detailed analysis

Write $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$ , $\tan\beta=\frac{\sin\beta}{\cos\beta}$ , $\tan\frac{\gamma}{2}=\frac{\sin(\gamma/2)}{\cos(\gamma/2)}$ in $a+b=\tan\frac{\gamma}{2}(a\tan\alpha+b\tan\beta)$ , then multiply both sides by $\cos\alpha\cos\beta\cos\frac{\gamma}{2}$ to clear every denominator.