Question

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

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

Answer 1

The correct option is C

$\frac{3}{4}$

$\angle C=2\frac{\pi }{3}\therefore A+B=\frac{\pi }{3}$
Now $co{s}^{2}A+co{s}^{2}B-cosAcosB=\frac{1+cos2A}{2}+\frac{1+cos2B}{2}-\frac{1}{2}\left[cos\right(A+B)+cos(A-B\left)\right]$
$=\frac{1}{2}[2+2cos(A+B\left)cos\right(A-B\left)\right]-\frac{1}{2}[\frac{1}{2}+cos(A-B\left)\right]=\frac{1}{2}[2+2\frac{1}{2}cos(A-B\left)\right]-\frac{1}{4}-\frac{1}{2}cos(A-B)=1-\frac{1}{4}=\frac{3}{4}$