Answer: see proof below
Step-by-step explanation:
Given: A + B + C = π → A + B = π - C
→ B + C = π - A
→ A + C = π - B
Use the following Double Angle Identity: sin 2A = 2 sin A · cos A
Use the following Cofunction Identity: sin A = cos (π/2 - A)
Use the following Sum to Product Identity:
sin A + sin B = sin [(A + B)/2] · cos [(A - B)/2]
Proof LHS → RHS
[tex]\text{LHS:}\qquad \dfrac{\cos A}{\sin B\cdot \sin C}+\dfrac{\cos B}{\sin C\cdot \sin A}+\dfrac{\cos C}{\sin A\cdot \sin B}\\\\\\.\qquad \quad = \bigg(\dfrac{2\sin A}{2\sin A}\bigg)\dfrac{\cos A}{\sin B\cdot \sin C}+\bigg(\dfrac{2\sin B}{2\sin B}\bigg)\dfrac{\cos B}{\sin C\cdot \sin A}+\bigg(\dfrac{2\sin C}{2\sin C}\bigg)\dfrac{\cos C}{\sin A\cdot \sin B}\\\\\\.\qquad \quad =\dfrac{2\sin A\cdot \cos A+2\sin B\cdot \cos B+2\sin C\cdot \cos C}{2\sin A\cdot \sin B\cdot \sin C}[/tex]
[tex]\text{Double Angle:}\qquad \qquad \dfrac{\sin 2A+\sin 2B+\sin 2C}{2\sin A\cdot \sin B\cdot \sin C}\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{(\sin 2A+\sin 2B)+\sin 2C}{2\sin A\cdot \sin B\cdot \sin C}[/tex]
[tex]\text{Sum to Product:}\qquad \dfrac{2\sin (A+B)\cdot \cos (A-B)+\sin 2C}{2\sin A\cdot \sin B\cdot \sin C}[/tex]
[tex]\text{Given:}\qquad \qquad \qquad \dfrac{2\sin (\pi -C)\cdot \cos (A-B)+\sin 2C}{2\sin A\cdot \sin B\cdot \sin C}\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{2\sin C\cdot \cos (A-B)+\sin 2C}{2\sin A\cdot \sin B\cdot \sin C}[/tex]
[tex]\text{Double Angle:}\qquad \qquad \dfrac{2\sin C\cdot \cos (A-B)+2\sin C\cdot \cos C}{2\sin A\cdot \sin B\cdot \sin C}\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{2\sin C(\cos (A-B)+\cos C)}{2\sin A\cdot \sin B\cdot \sin C}[/tex]
[tex]\text{Sum to Product:}\qquad \dfrac{2\sin C(2\cos (\frac{A-B+C}{2})\cdot \cos (\frac{A-B-C}{2})}{2\sin A\cdot \sin B\cdot \sin C}\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{4\sin C\cdot \cos (\frac{A+C}{2}-\frac{B}{2})\codt \cos (\frac{A}{2}-\frac{B+C}{2})}{2\sin A\cdot \sin B\cdot \sin C}[/tex]
[tex]\text{Given:}\qquad \qquad \dfrac{4\sin C(\frac{\pi-B}{2}-\frac{B}{2})\cdot \cos (\frac{A}{2}-\frac{\pi -A}{2})}{2\sin A\cdot \sin B\cdot \sin C}\\\\\\.\qquad \qquad \qquad =\dfrac{4\sin C(\frac{\pi}{2}-B)\cdot \cos (\frac{\pi}{2}-A)}{2\sin A\cdot \sin B\cdot \sin C}[/tex]
[tex]\text{Cofunction:}\qquad \qquad \dfrac{4\sin C\cdot \sin B\sin A}{2\sin A\cdot \sin B\cdot \sin C}\\\\\\.\qquad \qquad \qquad \qquad =2[/tex]
LHS = RHS: 2 = 2 [tex]\checkmark[/tex]
