# If (cosA+2cosC)/(cosA+2cosB)=(sinB)/sinC then show that the triangle is either isosceles or right angled triangle. I got cosA+cosB+cosC=1 what to do after this?

Jan 18, 2018

$\frac{\cos A + 2 \cos C}{\cos A + 2 \cos B} = \frac{\sin B}{\sin} C$

$\implies \left(\cos A + 2 \cos C\right) \sin C = \left(\cos A + 2 \cos B\right) \sin B$

$\implies \cos A \sin C + 2 \cos C \sin C = \cos A \sin B + 2 \cos B \sin B$
$\implies \cos A \sin C + \sin 2 C = \cos A \sin B + \sin 2 B = 0$
$\implies \cos A \left(\sin B - \sin C\right) + \left(\sin 2 B - \sin 2 C\right) = 0$

$\implies \cos A \cdot 2 \cos \left(\frac{B + C}{2}\right) \sin \left(\frac{B - C}{2}\right) + 2 \cos \left(B + C\right) \sin \left(B - C\right) = 0$

$\implies \cos A \cos \left(\frac{\pi}{2} - \frac{A}{2}\right) \sin \left(\frac{B - C}{2}\right) + \cos \left(\pi - A\right) \sin \left(B - C\right) = 0$

$\implies \cos A \sin \left(\frac{A}{2}\right) \sin \left(\frac{B - C}{2}\right) - \cos A \cdot 2 \sin \left(\frac{B - C}{2}\right) \cos \left(\frac{B - C}{2}\right) = 0$

$\implies \cos A \sin \left(\frac{B - C}{2}\right) \left(\sin \left(\frac{A}{2}\right) - 2 \cos \left(\frac{B - C}{2}\right)\right) = 0$

So either $\cos A = 0 = \cos \left(\frac{\pi}{2}\right)$

$\implies A = \frac{\pi}{2} \to$triangle is right angled.

Or $\sin \left(\frac{B - C}{2}\right) = 0$

$\implies B = C \to$ the triangle is isoscles