Question

If in a triangle $A B C, \angle C=\frac{2 \pi}{3}$, 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)

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