In any $\triangle A B C, a(b \cos C-c \cos B)$ equals

In any $\triangle A B C, a(b \cos C-c \cos B)$ equals
  1. $b^2+c^2$
  2. $b^2-c^2$
  3. $\frac{1}{b}+\frac{1}{c}$
  4. $\frac{1}{b^2}-\frac{1}{c^2}$

Solution

$\begin{aligned} a(b \cos C-c \cos B) & =a b \cos C-a c \cos B \\ & =\frac{a^2+b^2-c^2}{2}-\frac{a^2+c^2-b^2}{2} \\ & =b^2-c^2\end{aligned}$

Asked in: AP EAMCET 2009

Practice more Trigonometric Functions questions on Aicharya