Question

Evaluate: $\frac{{\cos }^{2}20+{\cos }^{2}70}{{\sin }^{2}59+{\sin }^{2}31}$

Answer 1

we know that $\cos ({90}^{\circ }-\theta )=\sin \theta$

$\sin ({90}^{\circ }-\theta )=\cos \theta$

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

$=\frac{{\cos }^2{20}^{\circ }+{\sin }^2{20}^{\circ }}{{\sin }^2{59}^{\circ }+{\cos }^2{59}^{\circ }}$

$=\frac{1}{1}$ (as ${\cos }^2\theta +{\sin }^2\theta =1)$

$=1$