In any triangle, if $\frac {\cos A+2\cos C}{\cos A+2\cos B}=\frac {\sin B}{\sin C}$, then the triangle is either isosceles or right-angled In any triangle, if $\dfrac {\cos A+2\cos C}{\cos A+2\cos B}=\dfrac {\sin B}{\sin C}$, prove that the triangle is either isosceles or right angled.
My Attempt:
Given:
$$\dfrac {\cos A+2\cos C}{\cos A+2\cos B}=\dfrac {\sin B}{\sin C}$$
$$\dfrac {\dfrac {b^2+c^2-a^2}{2bc}+2\dfrac {a^2+b^2-c^2}{2ab}}{\dfrac {b^2+c^2-a^2}{2bc}+2\dfrac {a^2+c^2-b^2}{2ac}}=\dfrac {b}{c}$$
On simplification,
$$\dfrac {ab^2+ac^2-a^3+2a^2c+2b^2c-2c^3}{ab^2+ac^2-a^3+2a^2b+2bc^2-2b^3}=\dfrac {b}{c}$$
 A: In your work try to factor out a term like $(b-c)$ from the expression:
$$c(ab^2+ac^2-a^3+2a^2c+2b^2c-2c^3) = b(ab^2+ac^2-a^3+2a^2b+2bc^2-2b^3)\\
ab^2c+ac^3-a^3c + 2a^2c^2 +2b^2c^2 - 2 c^4 = ab^3 + abc^2 -a^3b + 2a^2 b^2 + 2b^2 c^2 -2b^4\\
abc(b-c) + a(c^3 - b^3 ) +a^3 (b-c)+2a^2 (c^2 - b^2) +2(b^4-c^4)$$
From here either $b=c$, or
$$abc - a(c^2+cb+b^2)+a^3 - 2a^2(c+b) + 2(b^3 + b^2c+bc^2 +c^3)=0\\
- ac^2 -ab^2 +a^3 -2a^2c - 2a^2 b + 2b^3 + 2b^2c+2bc^2 + 2c^3 = 0 \\
-2c(a^2-b^2 -c^2)+a(a^2-b^2-c^2)-2b(a^2-b^2-c^2) = 0\\
(a-2b-2c)(a^2-b^2-c^2)=0$$
This is only possible if $a^2 = b^2 +c^2 $ as $a-2b-2c$ is always negative due to triangle inequality.
A: Alternate approach:
Cross multiply
$\cos A \sin C + \sin 2C = \sin B \cos A + \sin 2B$
$\cos A (\sin C - \sin B) - (\sin 2B - \sin 2C)=0$
$\cos A * 2* \sin \frac{C-B}{2} \cos \frac{B+C}{2} - 2* \sin (B-C) \cos(B+C)=0$
$\cos A * 2* \sin \frac{C-B}{2} \cos \frac{B+C}{2} + 2* \sin (B-C) \cos A=0$
$2\cos A [\sin \frac{C-B}{2} \cos \frac{B+C}{2} + 2 \sin \frac{C-B}{2} \cos \frac{B-C}{2}] = 0$
$2\cos A *\sin \frac{B-C}{2}[\cos \frac{B+C}{2} - 2\cos\frac{B-C}{2}]=0$ 
IF:
$\cos \frac{B+C}{2} - 2\cos\frac{B-C}{2} =0 \implies \tan \frac B2 \tan \frac C2 = -\frac 13$ and $\frac B2<90 $ and $\frac C2 < 90$ 
A: You are very close to answer. 


*

*cross multiply.

*subtract 1 from both side.

*take b-c common from Numerator.

*It's done!! :)
