Question

If $\cos A+\cos C=4{\sin }^2\frac{1}{2}B$, the sides $a,b,c$ of the triangle are in $G.P.$

• A. True
• B. False

Answer 1

The correct option is A True

$\cos A+\cos C=4{\sin }^2\frac{B}{2}$

$\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+b^2-c^2}{2ab}=4\frac{(s-c)(s-a)}{ca}$

$\Rightarrow \frac{(b^2+ac)(a+c)-(a^3+c^3)}{2abc}\Rightarrow \frac{4(a+b-c)(b+c-a)}{\left(4\right)\left(ca\right)}$

$\Rightarrow (b^2+ac)(a+c)-(a^3+c^3)$

$\Rightarrow \frac{8b}{4}(ab+ac-a^2+b^2+bc-ab-bc-c^2+ac)$

$\Rightarrow (b^2+ac)(a+c)-(a^2+c^3)$

$\Rightarrow \frac{8b}{4}(-a^4+b^2-c^2+2ac)$

$\Rightarrow (b^2+ac)(a+c)-(a+c)(a^2+c^2-ac)$

$\Rightarrow 2b(b^2-(a-c{)}^2)$

$\Rightarrow (a+c)(b^2-(a^2+c^2-2ac\left)\right)$

$\Rightarrow 2b(b^2-(a-c{)}^2)$

$a+c=2b$

They are in $A.P$