Question

Without using tables, evaluate :
$\frac{{\cos }^2{20}^{\circ }+{cos}^2{70}^{\circ }}{{sin}^2{59}^{\circ }+{sin}^2{31}^{\circ }}$

Answer 1

Given $\frac{{\cos }^2{20}^{\circ }+{\cos }^2{70}^{\circ }}{{\sin }^2{59}^{\circ }+{\sin }^2{31}^{\circ }}$

$=\frac{{\cos }^2({90}^{\circ }-{70}^{\circ })+{\cos }^2{70}^{\circ }}{{\sin }^2({90}^{\circ }-{31}^{\circ })+{\sin }^2{31}^{\circ }}$
Since, $\sin (90-x{)}^{\circ }=\cos x^{\circ }$ and $\cos (90-x{)}^{\circ }=\sin x^{\circ }$

$=\frac{{\sin }^2{70}^{\circ }+{\cos }^2{70}^{\circ }}{{\cos }^2{31}^{\circ }+{\sin }^2{31}^{\circ }}$
Since, ${\sin }^{2}x+{\cos }^{2}x=1$

$\Rightarrow \frac{{\sin }^2{70}^{\circ }+{\cos }^2{70}^{\circ }}{{\cos }^2{31}^{\circ }+{\sin }^2{31}^{\circ }}=\frac{1}{1}$
Therefore, the value of the given expression is $1$.