This question was uploaded on 08/11/21 on social media accounts.
This is a standard case where if the tangent of half of the angles of triangles is in A.P then the cosine of those angles are also in A.P.
Solution:
Given that:
\[\alpha+\beta+\gamma=180\]
\[\Rightarrow \alpha+\gamma=180-\beta\]
Also given that:
\[\tan\frac{\alpha}2+\tan\frac{\gamma}2 = 2\tan\frac{\beta}2\]
These three terms are in A.P.
\[\Rightarrow \tan\frac{\alpha}2-\tan\frac{\beta}2 = \tan\frac{\beta}2-\tan\frac{\gamma}2\]
\[\Rightarrow \frac{\sin\frac{\alpha}2}{\cos\frac{\alpha}2}-\frac{\sin\frac{\beta}2}{\cos\frac{\beta}2} = \frac{\sin\frac{\beta}2}{\cos\frac{\beta}2}-\frac{\sin\frac{\gamma}2}{\cos\frac{\gamma}2}\]
\[\Rightarrow \frac{\sin\frac{\alpha}2\cos\frac{\beta}2-\cos\frac{\alpha}2\sin\frac{\beta}2}{\cos\frac{\alpha}2\cos\frac{\beta}2} = \frac{\sin\frac{\beta}2\cos\frac{\gamma}2-\cos\frac{\beta}2\sin\frac{\gamma}2}{\cos\frac{\beta}2\cos\frac{\gamma}2}\]
\[\Rightarrow \frac{\sin\frac{\alpha-\beta}2}{\cos\frac{\alpha}2} = \frac{\sin\frac{\beta-\gamma}2}{\cos\frac{\gamma}2}\]
\[\Rightarrow \frac{\sin\frac{\alpha-\beta}2}{\sin\frac{\beta+\gamma}2} = \frac{\sin\frac{\beta-\gamma}2}{\sin\frac{\alpha+\beta}2}\]
\[\Rightarrow \sin\frac{\alpha-\beta}2\sin\frac{\alpha+\beta}2 = \sin\frac{\beta-\gamma}2\sin\frac{\beta+\gamma}2\]
\[\Rightarrow \cos\alpha-\cos\beta = \cos\beta-\cos\gamma\]
That means, cos terms are in A.P.
\[\Rightarrow \cos\alpha+\cos\gamma = 2\cos\beta\]
\[\Rightarrow \frac{\cos\alpha+\cos\gamma}{\cos\beta}=2\]
