Question

Find the value of $[\frac{{\sin }^2{22}^o+{\sin }^2{68}^o}{{\cos }^2{22}^o{\cos }^2{68}^o}+{\sin }^3{63}^o+\cos {63}^o\sin {27}^o]$

Answer 1

$[\frac{si{n}^2{22}^{\circ }+si{n}^2{68}^{\circ }}{cose{c}^{2}22\circ co{s}^2{68}^{\circ }}+si{n}^2{63}^{\circ }+cos{63}^{\circ }sin{27}^{\circ }]$

$[\frac{si{n}^2{22}^{\circ }+si{n}^2({90}^{\circ }-{22}^{\circ })}{cose{c}^2({90}^{\circ }-{68}^{\circ })co{s}^2{68}^{\circ }}+si{n}^2{63}^{\circ }+cos{63}^{\circ }sin({90}^{\circ }-{63}^{\circ }\left)\right]$

$[\frac{si{n}^2{22}^{\circ }+co{s}^2{22}^{\circ }}{se{c}^2{68}^{\circ }co{s}^2{68}^{\circ }}+si{n}^2{63}^{\circ }+cos{63}^{\circ }cos{63}^{\circ }]$

$\frac{1}{sec\theta }=cos\theta$ $and$ $si{n}^2\theta +co{s}^2\theta =1$

$[\frac{1}{1}+si{n}^2{63}^{\circ }+co{s}^2{63}^{\circ }]$

$1+1$

$=2$