Question

The value of ${\cos }^2{10}^{}-\cos {10}^{}\cos {50}^{}+{\cos }^2{50}^{}$ is equal to

• A. $\frac{3}{4}$
• B. $\frac{1}{3}$
• C. $\frac{3}{2}$
• D. $0$

Answer 1

The correct option is A $\frac{3}{4}$

${\cos }^2{10}^o-\cos {10}^o.\cos {50}^o+{\cos }^2{50}^o$

$={\cos }^2{50}^o-{\sin }^2{10}^o+1-(2.\cos {50}^o.\cos {10}^o).\frac{1}{2}$

$=\cos {60}^o.\cos {40}^o+1-\frac{1}{2}\left(\cos \right({60}^o)+\cos ({40}^o\left)\right)$

$(\because {\cos }^{2}A-{\sin }^{2}B=\cos (A+B).\cos (A-B)\text{and}2\cos A.\cos B=\cos (A+B)+\cos (A-B\left)\right)$

$=\frac{1}{2}\cos {40}^o+1-\frac{1}{2}(\frac{1}{2}+\cos {40}^o)$

$=1-\frac{1}{4}$

$=\frac{3}{4}$