Question

$\frac{2{\sin }^{2}63°+1+2{\sin }^{2}27°}{3{\cos }^{2}17°-2+3{\cos }^{2}73°}=?$

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

(b) $\frac{3}{2}$

(c) 2

(d) 3

Answer 1

$$\begin{gathered} \frac{2{\sin }^{2}63°+1+2{\sin }^{2}27°}{3{\cos }^{2}17°-2+3{\cos }^{2}73°} \\ =\frac{2{\left(\sin \left(90°-27°\right)\right)}^2+1+2{\sin }^{2}27°}{3{\left(\cos \left(90°-73°\right)\right)}^2-2+3{\cos }^{2}73°} \\ =\frac{2{\cos }^{2}27°+1+2{\sin }^{2}27°}{3{\sin }^{2}73°-2+3{\cos }^{2}73°}\left(\because \sin \left(90°-\theta \right)=\cos \theta \mathrm{and}\cos \left(90°-\theta \right)=\sin \theta \right) \\ =\frac{2\left({\cos }^{2}27°+{\sin }^{2}27°\right)+1}{3\left({\sin }^{2}73°+{\cos }^{2}73°\right)-2} \\ =\frac{2\left(1\right)+1}{3\left(1\right)-2}\left(\mathrm{using}\mathrm{the}\mathrm{identity}:\hspace{0.17em}{\sin }^2\theta +{\cos }^2\theta =1\right) \\ =\frac{2+1}{3-2} \\ =3 \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{d}\right). \end{gathered}$$