Question

$\frac{2{\sin }^{2}63°+1+2{\sin }^{2}27°}{3{\cos }^{2}17°-2+3{\cos }^{2}73°}=?$
(a) $\frac{3}{2}$
(b) $\frac{2}{3}$
(c) 2
(d) 3

Answer 1

$$\begin{gathered} \left(\mathrm{d}\right)3 \\ \text{Given:}\frac{2{\sin }^2{63}^0+1+2{\sin }^2{27}^0}{3{\cos }^2{17}^0-2+3{\cos }^2{73}^0} \\ =\frac{2({\sin }^2{63}^0+{\sin }^2{27}^0)+1}{3({\cos }^2{17}^0+{\cos }^2{73}^0)-2} \\ =\frac{2[{\sin }^2{63}^0+{\sin }^2({90}^0-{63}^0)]+1}{3[{\cos }^2{17}^0+{\cos }^2({90}^0-{17}^0)]-2} \\ =\frac{2({\sin }^2{63}^0+{\cos }^2{63}^0)+1}{3({\cos }^2{17}^0+{\sin }^2{17}^0)-2}[\because \sin ({90}^0-\theta )=\cos \theta \text{and}\cos ({90}^0-\theta )=\sin \theta ] \\ =\frac{2\times 1+1}{3\times 1-2}[\because {\sin }^2\theta +{\cos }^2\theta =1] \\ =\frac{2+1}{3-2} \\ =\frac{3}{1}=3 \end{gathered}$$