# Solve this

Question:

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

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

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

(c) 2

(d) 3

Solution:

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

$=\frac{2\left(\sin \left(90^{\circ}-27^{\circ}\right)\right)^{2}+1+2 \sin ^{2} 27^{\circ}}{3\left(\cos \left(90^{\circ}-73^{\circ}\right)\right)^{2}-2+3 \cos ^{2} 73^{\circ}}$

$=\frac{2 \cos ^{2} 27^{\circ}+1+2 \sin ^{2} 27^{\circ}}{3 \sin ^{2} 73^{\circ}-2+3 \cos ^{2} 73^{\circ}} \quad\left(\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right.$ and $\left.\cos \left(90^{\circ}-\theta\right)=\sin \theta\right)$

$=\frac{2\left(\cos ^{2} 27^{\circ}+\sin ^{2} 27^{\circ}\right)+1}{3\left(\sin ^{2} 73^{\circ}+\cos ^{2} 73^{\circ}\right)-2}$

$=\frac{2(1)+1}{3(1)-2} \quad$ (using the identity : $\sin ^{2} \theta+\cos ^{2} \theta=1$ )

$=\frac{2+1}{3-2}$

$=3$

Hence, the correct option is (d).

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