This question was uploaded on 19/11/21 on social media accounts.
Solution:
Given that A+B+C=π
Let us focus on: sinA + sinB + sinC
\[\sin A + \sin B + \sin C\]
\[= 2\sin\left(\frac{A+B}2\right)\cos\left(\frac{A-B}2\right) + 2\sin\frac C2\cos\frac C2\]
A+B=π-C
\[= 2\sin\left(\frac\pi2 + \frac C2\right)\cos\left(\frac{A-B}2\right) + 2\sin\frac C2\cos\frac C2\]
\[= 2\cos\frac C2 \cos\left(\frac{A-B}2\right) + 2\sin\frac C2\cos\frac C2\]
\[= 2\cos\frac C2 \left[\cos\left(\frac{A-B}2\right) + \sin\frac C2\right]\]
C=π-(A+B)
\[= 2\cos\frac C2 \left[\cos\left(\frac{A-B}2\right) + \sin \left(\frac\pi2- \frac{A+B}2\right)\right]\]
\[= 2\cos\frac C2 \left[\cos\left(\frac{A-B}2\right) + \cos \left(\frac{A+B}2\right)\right]\]
\[= 2\cos\frac C2 \left[2\cos\frac{A}2\cos\frac{B}2\right]\]
\[= 4\cos\frac A2\cos\frac B2\cos\frac C2\]
\[\color{blue} {\Rightarrow \sin A + \sin B + \sin C= 4\cos\frac A2\cos\frac B2\cos\frac C2}\]
By putting this in our fraction then all terms will directly cancel.
\[\Rightarrow \frac{\sin A + \sin B + \sin C} {\cos\frac A2 \cdot \cos\frac B2 \cdot \cos\frac C2} = 4\]
