Question

The value of $\frac{{\cos }^2{20}^o+{\cos }^2{70}^o}{{\sin }^2{59}^o+{\sin }^2{31}^o}$ is :

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is C $1$

${\cos }^2{20}^o+{\cos }^2{70}^o$
$={\cos }^2({90}^o-{70}^o)+{\cos }^2{70}^o$
$={\sin }^2{70}^o+{cos}^2{70}^o=1$ .... $[\because \cos (90-\theta )=\sin \theta ]$

${\sin }^2{59}^o+{\sin }^2{31}^o={\sin }^2{59}^o+{\sin }^2({90}^o-{59}^o)$
$={\sin }^2{59}^o+{\cos }^2{59}^o=1$

$\therefore \frac{{\cos }^2{20}^o+{\cos }^2{70}^o}{{\sin }^2{59}^o+{\sin }^2{31}^o}=\frac{1}{1}=1$