Question : The value of $\frac{\sin ^2 52^{\circ}+2+\sin ^2 38^{\circ}}{4 \cos ^2 43^{\circ}-5+4 \cos ^2 47^{\circ}}$ is:
Option 1: $3$
Option 2: $\frac{1}{3}$
Option 3: $-\frac{1}{3}$
Option 4: $-3$
Correct Answer: $-3$
Solution :
Given: $\frac{\sin ^2 52^{\circ}+2+\sin ^2 38^{\circ}}{4 \cos ^2 43^{\circ}-5+4 \cos ^2 47^{\circ}}$
= $\frac{\sin ^2 52^{\circ}+2+\sin ^2 (90^{\circ} - 52^{\circ})}{4 \cos ^2 43^{\circ}-5+4 \cos ^2 (90^{\circ} - 43^{\circ})}$
= $\frac{\sin ^2 52^{\circ}+2+\cos ^2 52^{\circ}}{4 \cos ^2 43^{\circ}-5+4 \sin ^2 43^{\circ}}$
= $\frac{1+2}{4 (\cos ^2 43^{\circ}+ \sin ^2 43^{\circ})-5}$
= $\frac{1+2}{4-5}$
= $\frac{3}{-1}$
= $-3$
Hence, the correct answer is $-3$.
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