Question

$({\cos }^2{37}^{\circ }-{\sin }^2{53}^{\circ })=$

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

Answer 1

The correct option is D $0$

Consider given term

$({\cos }^2{37}^o-{\sin }^2{53}^o)$

$={\cos }^2{37}^o-{\sin }^2\left({90}^o-{37}^o\right)$

We know that $\sin (90-\theta )=cos\theta$

$={\cos }^2{37}^o-{\cos }^2{37}^o$

$=0$