Question

If in a triangle $ABC,\angle C=\frac{2\pi }{3}$, then the value of ${\cos }^{2}A+{\cos }^{2}B-\cos A\cdot \cos B$ is equal to-

• A. $\frac{3}{4}$
• B. $\frac{3}{2}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$A+B=\frac{\pi }{3}$
$\therefore \cos (A+B)=\frac{1}{2}$
${\cos }^{2}A+{\cos }^{2}B-\cos A\cos B$
$={\cos }^{2}A-{\sin }^{2}B+1-\cos A\cos B$
$=\cos (A+B)\cos (A-B)+1-\cos A\cos B$
$=\frac{1}{2}(\cos A\cos B+\sin A\sin B)+1-\cos A\cos B$
$=1-\frac{1}{2}(\cos A\cos B-\sin A\sin B)$
$=1-\frac{1}{2}\cos (A+B)=\frac{3}{4}$