Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively,

then,$\frac{a\cos C+c\cos A}{\cos B}$ is equal to

• A. $\frac{2abc}{a^2+b^2+c^2}$
• B. $\frac{abc}{b^2+c^2-a^2}$
• C. $\frac{abc}{b^2+c^2+a^2}$
• D. $\frac{2abc}{a^2+c^2-b^2}$

Answer 1

The correct option is D $\frac{2abc}{a^2+c^2-b^2}$

Given a $\Delta$ ABC whose sides are a,b,c & angles are A,B,C

By $\text{cosine formula}$,

$cos\left(B\right)=\frac{a^2+c^2-b^2}{2ac}$

By $\text{Projection formula}$,

$(a\times \cos C+c\times \cos A)=b$

Therefore, $\frac{a\left(\cos C\right)+c\left(\cos A\right)}{\cos B}=\frac{b}{\cos B}$

$=\frac{b\times \left(2ac\right)}{a^2+c^2-b^2}$

$=\frac{2abc}{a^2+b^2-c^2}$

hence, option (D) is correct.