Question

If $A+B=C$, then ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C-2\cos A\cos B\cos C=$

• A. $1$
• B. $2$
• C. $0$
• D. $3$

Answer 1

The correct option is A $1$

Let

$I={\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C-2\cos A\cos B\cos C$

$=\frac{1}{2}(\cos 2A+1+\cos 2B+1+cos2C+1)-2\cos A\cos B\cos C$

$=\frac{1}{2}(\cos 2A+\cos 2B+\cos 2C+3)-2\cos A\cos B\cos C$

Using

$C=A+B$

$I=\frac{1}{2}\left(2\cos \right(A+B\left)\cos \right(A-B)+\cos 2((A+B))+3)-2\cos A\cos B\cos C$

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

$=\frac{1}{2}\left(2\cos \right(A+B\left)\right(\cos (A-B)+\cos (A+B))+2)-2\cos A\cos B\cos C$

Again using

$A+B=C$

$I=\frac{1}{2}\left(2\cos \right(C\left)\right(2\cos A\cos B)+2)-2\cos A\cos B\cos C$

$=2\cos A\cos B\cos C+1-2\cos A\cos B\cos C=1$