Question

Solve:
$\frac{{\sin }^2{63}^o+{\sin }^2{27}^o}{{\cos }^2{17}^o+{\cos }^2{73}^o}$

Answer 1

$\frac{si{n}^2{63}^{\circ }+si{n}^2{27}^{\circ }}{co{s}^2{17}^{\circ }+co{s}^2{73}^{\circ }}$

we know sin $(90-\theta )=cos\theta \&cos(90-\theta )=sin\theta$ .

$\Rightarrow \frac{si{n}^{2}63+si{n}^2(90-27)}{co{s}^{2}17+si{n}^2(90-73)}$ =$\frac{si{n}^{2}63+co{s}^2\left(63\right)}{co{s}^{2}17+si{n}^{2}17}$

$co{s}^2\theta +si{n}^2\theta =1$

$\Rightarrow \frac{1}{1}$ =1