I saw this problem in a math magazine:
Let $\alpha,\beta$ and $\gamma$ be the angles of a triangle, so that
$$
\alpha+\beta+\gamma=\pi.
$$
Then it holds that
$$
\sum_{cyc} \frac{\cos(\frac{\alpha+\beta}{2})}{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}}=\frac{\cos\left(\frac{\alpha+\beta}{2}\right)}{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}}+\frac{\cos\left(\frac{\alpha+\gamma}{2}\right)}{\cos \frac{\alpha}{2}\cos \frac{\gamma}{2}}+
\frac{\cos\left(\frac{\gamma+\beta}{2}\right)}{\cos \frac{\gamma}{2}\cos \frac{\beta}{2}}=2.
$$
I tried to use $\alpha+\beta=\pi-\gamma$ so that
$$
\cos\left(\frac{\alpha+\beta}{2}\right)=\cos\left(\frac{\pi}{2}-\frac{\gamma}{2}\right)= \cos \frac{\pi}{2}\cos \frac{\gamma}{2}+ \sin\frac{\pi}{2}\sin \frac{\gamma}{2}= \sin\frac{\gamma}{2}.
$$
Now the above equation equals
$$
\frac{\sin\frac{\gamma}{2}}{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}}+\frac{\sin\frac{\beta}{2}}{\cos \frac{\alpha}{2}\cos \frac{\gamma}{2}}+
\frac{\sin\frac{\alpha}{2}}{\cos \frac{\gamma}{2}\cos \frac{\beta}{2}}=2.
$$
But now I don't have ideas for continuation. Any hints to tackle this?
Best Answer
First off the bat, your formula in the first line is wrong, it should be, as per my comment, $$\cos\left(\frac{\pi}{2}-\frac{\gamma}{2}\right) = \cos \frac{\pi}{2}\cos \frac{\gamma}{2}\color{purple}+\sin\frac{\pi}{2}\sin \frac{\gamma}{2}=\sin\frac{\gamma}{2}$$
I’ll take an alternative route. Expand $\cos\left(\dfrac{\alpha+\beta}{2}\right)$ as $\cos\dfrac{\alpha}{2}\cos\dfrac{\beta}{2}-\sin \dfrac{\alpha}{2}\sin \dfrac{\beta}{2}$ so that $$\dfrac{\cos\left(\dfrac{\alpha+\beta}{2}\right)}{\cos\dfrac{\alpha}{2}\cos\dfrac{\beta}{2}}=\dfrac{\cos\dfrac{\alpha}{2}\cos\dfrac{\beta}{2}-\sin \dfrac{\alpha}{2}\sin \dfrac{\beta}{2}}{\cos\dfrac{\alpha}{2}\cos\dfrac{\beta}{2}}$$$$=1-\tan\frac{\alpha}{2}\tan\frac{\beta}{2}$$
So summing similar terms we get $$LHS=3-\left(\tan\frac{\alpha}{2}\tan\frac{\beta}{2}+ \tan\frac{\gamma}{2}\tan\frac{\beta}{2}+ \tan\frac{\gamma}{2}\tan\frac{\beta}{2}\right)\tag{1}$$
Now, $$\alpha+\beta+\gamma=\pi$$$$\implies\frac{\alpha}{2}+\frac{\beta}{2}=\frac{\pi}{2}-\frac{\gamma}{2}$$ so that taking tangents and expanding using the formula gives us $$\frac{\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}}{1-\tan\frac{\alpha}{2}\tan\frac{\beta}{2}}=\tan\left(\frac{\pi}{2}-\frac{\gamma}{2}\right)=\frac{1}{\tan\frac{\gamma}{2}}.\tag{2}$$ Rearranging gives us $$ \tan\frac{\alpha}{2}\tan\frac{\beta}{2}+ \tan\frac{\gamma}{2}\tan\frac{\beta}{2}+ \tan\frac{\gamma}{2}\tan\frac{\beta}{2}=1$$ so using $(1)$ and $(2)$, we get our desired result.