Question

In $△ABC$ if $A,B$ and $C$ represent the angles of a triangle, then the value of ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C-1$ is equal to

• A. $-\cos A\cos B\cos C$
• B. $-2\cos A\cos B\cos C$
• C. $2\cos A\cos B\cos C$
• D. $\cos A\cos B\cos C$

Answer 1

The correct option is B $-2\cos A\cos B\cos C$

${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C-1=\frac{1+\cos 2A}{2}+\frac{1+\cos 2B}{2}+\frac{1+\cos 2C}{2}-1=\frac{\cos 2A+\cos 2B+\cos 2C}{2}+\frac{3}{2}-1=\frac{2\cos (A+B)\cos (A-B)+2{\cos }^{2}C-1}{2}+\frac{1}{2}=\frac{2\cos (A+B)\cos (A-B)+2{\cos }^{2}C}{2}=\frac{2\cos (A+B)\cos (A-B)+2{\cos }^2(\pi -(A+B\left)\right)}{2}(\because A+B+C=\pi )=\frac{2\cos (A+B)\left[\cos \right(A-B)+\cos (A+B)}{2}=\frac{2\cos (\pi -C)\left[2\cos A\cos B\right]}{2}=-2\cos A\cos B\cos C$