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 2 of 6: Group by

a

and

b

using angle addition

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In plain words

Recognizing $\cos\theta\cos\phi-\sin\theta\sin\phi$ as $\cos(\theta+\phi)$ collapses four trig terms into two clean cosine factors.

a\cos\beta\cos\left(\alpha+\frac{\gamma}{2}\right) + b\cos\alpha\cos\left(\beta+\frac{\gamma}{2}\right) = 0

Detailed analysis

Move every term to one side and factor: the $a$ -terms combine as $a\cos\beta(\cos\alpha\cos\frac{\gamma}{2}-\sin\alpha\sin\frac{\gamma}{2})$ and the $b$ -terms as $b\cos\alpha(\cos\beta\cos\frac{\gamma}{2}-\sin\beta\sin\frac{\gamma}{2})$ . Each parenthesis equals $\cos(\theta+\frac{\gamma}{2})$ by the cosine addition formula, giving the stated identity.