Question

Evaluate : $\frac{{\sin }^2{63}^{\circ }+{\sin }^2{27}^{\circ }}{{\cos }^2{17}^{\circ }+{\cos }^2{73}^{\circ }}$.

Answer 1

Given

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

[we know that $63+27={90}^{\circ }]$

Hence ${63}^{\circ }={90}^{\circ }-{27}^{\circ }$

similarly ${17}^{\circ }+{73}^{\circ }={90}^{\circ }$

Hence ${17}^{\circ }={90}^{\circ }-{73}^{\circ }]$

$=\frac{si{n}^2({90}^{\circ }-{27}^{\circ })+si{n}^2{27}^{\circ }}{co{s}^2({90}^{\circ }-{73}^{\circ })+co{s}^2{73}^{\circ }}$

$[\therefore cos(90-\theta )=sin\theta ;sin(90-\theta )=cos\theta ]$

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

$[\therefore si{n}^{2}A+co{s}^{2}A=1]]$

$=\frac{1}{1}$

$=1.$

$\therefore \frac{si{n}^{2}63+co{s}^2{27}^{\circ }}{co{s}^{\circ }{77}^{\circ }+co{s}^2{73}^{\circ }}=1$