Question

In $\Delta ABC,\angle C=\frac{2\pi }{3}$, then the value of ${\cos }^{2}A+{\cos }^{2}B-\cos A.\cos B=$___

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

Answer 1

The correct option is A

$\frac{3}{4}$

In $\Delta ABC,\angle C=\frac{2\pi }{3}\therefore A+B=\frac{\pi }{3}$
Now ${\cos }^{2}A+{\cos }^{2}B-\cos A\cos B=\frac{1+\cos 2A}{2}+\frac{1+\cos 2B}{2}-\frac{1}{2}\left[\cos \right(A+B)+\cos (A-B\left)\right]$
$=\frac{1}{2}[2+2\cos (A+B\left)\cos \right(A-B\left)\right]-\frac{1}{2}[\frac{1}{2}+\cos (A-B\left)\right]=\frac{1}{2}[2+\frac{2\times 1}{2}\cos (A-B\left)\right]-\frac{1}{4}-\frac{1}{2}\cos (A-B)=1-\frac{1}{4}=\frac{3}{4}$

Hence the correct answer is Option C.