Question : What is the value of $\sec^2 54^{\circ}-\cot 2 36^{\circ}+\frac{3}{2} \sin^2 37^{\circ} \times \sec^2 53^{\circ}+\frac{2}{\sqrt{3}} \tan 60^{\circ}$?
Option 1: $\frac{5}{2}$
Option 2: $\frac{9}{2}$
Option 3: $\frac{3}{2}$
Option 4: $\frac{7}{2}$
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Correct Answer: $\frac{9}{2}$
Solution :
Given: $\sec^2 54^{\circ}-\cot ^2 36^{\circ}+\frac{3}{2}\sin^2 37^{\circ}\times \sec^2 53^{\circ}+\frac{2}{\sqrt{3}}\tan60^{\circ}$
$=\sec^2 (90^{\circ}-36^{\circ})-\cot^2 36^{\circ}+\frac{3}{2}\sin^2 (90^{\circ}-53^{\circ})\times \sec^2 53^{\circ}+\frac{2}{\sqrt{3}}×\sqrt3$
$=\operatorname{cosec}^2 36^{\circ}-\cot^2 36^{\circ}+\frac{3}{2}\cos^2 53^{\circ}\times \sec^2 53^{\circ}+2$
$=1+\frac{3}{2}\cos^2 53^{\circ}\times \frac{1}{\cos^2 53^{\circ}}+2$ ($\because \operatorname{cosec}^2 \theta-\cot ^2 \theta = 1$ )
$=1+\frac{3}{2}+2$
$=\frac{2+3+4}{2}$
$=\frac{9}{2}$
Hence, the correct answer is $\frac{9}{2}$.
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