Question

$\frac{2si{n}^2{63}^o+1+2si{n}^2{27}^o}{3co{s}^2{17}^o-2+3co{s}^2{73}^o}$ = ?

(a) $\frac{3}{2}$ (a) $\frac{2}{3}$ (c) 2 (d) 3

Answer 1

$\frac{2si{n}^{2}63°+1+2si{n}^{2}27°}{3co{s}^{2}17°-2+3co{s}^{2}73°}$

Here is the solution.

$\frac{2si{n}^{2}63°+1+2si{n}^{2}27°}{3co{s}^{2}17°-2+3co{s}^{2}73°}$
=$\frac{2[si{n}^{2}63°+si{n}^{2}27°]+1}{3[co{s}^{2}17°+co{s}^{2}73°]-2}$
=$\frac{2\left[si{n}^2\right(90°-27°)+si{n}^{2}27°]+1}{3\left[co{s}^2\right(90°-73°)+co{s}^{2}73°]-2}$
=$\frac{2[co{s}^{2}27+si{n}^{2}27°]+1}{3[si{n}^{2}73°+co{s}^{2}73°]-2}$ [as, sin(90°−θ)=cos θ; cos(90°−θ)=sin θ]

=$\frac{2\times 1+1}{3\times 1-2}$ [as, $si{n}^2\theta +co{s}^2\theta =1$]
= 3

correct option is (d) 3