Question

Question 11
The value of the expression
$(\frac{si{n}^2{22}^{\circ }+si{n}^2{68}^{\circ }}{co{s}^2{22}^{\circ }+co{s}^2{68}^{\circ }}+si{n}^2{63}^{\circ }+cos{63}^{\circ }{27}^{\circ })$ is

(A) 3
(B) 2
(C) 1
(D) 0

Answer 1

The answer is (B).

Given expression, $\frac{si{n}^2{22}^{\circ }+si{n}^2{68}^{\circ }}{co{s}^2{22}^{\circ }+co{s}^2{68}^{\circ }}+si{n}^2{63}^{\circ }+cos{63}^{\circ }sin{27}^{\circ }=\frac{si{n}^2{22}^{\circ }+si{n}^2({90}^{\circ }-{22}^{\circ })}{co{s}^2({90}^{\circ }-{68}^{\circ })+co{s}^2{68}^{\circ }}+si{n}^2{63}^{\circ }+cos{63}^{\circ }sin{27}^{\circ }=\frac{si{n}^2{22}^{\circ }+co{s}^2{22}^{\circ }}{si{n}^2{68}^{\circ }+co{s}^2{68}^{\circ }}+si{n}^2{63}^{\circ }+cos{63}^{\circ }cos{63}^{\circ }\left[\begin{array}{c}\because sin({90}^{\circ }-\theta )=cos\theta \\ andcos({90}^{\circ }-\theta )=sin\theta \end{array}\right]=\frac{1}{1}+(si{n}^2{63}^{\circ }+co{s}^2{63}^{\circ })[\because si{n}^2\theta +co{s}^2\theta =1]=1+1=2$