Question : The value of $\frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\tan ^2 60^{\circ}-\sin ^2 30^{\circ}-\cos ^2 45^{\circ}}$ is:
Option 1: $\frac{67}{27}$
Option 2: $\frac{22}{9}$
Option 3: $\frac{67}{24}$
Option 4: $\frac{19}{9}$
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Correct Answer: $\frac{67}{27}$
Solution :
Given expression,
$\frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\tan ^2 60^{\circ}-\sin ^2 30^{\circ}-\cos ^2 45^{\circ}}$
We know, $\cos 60^{\circ}=\frac12$, $\sec30^{\circ}=\frac2{\sqrt3}$, $\tan 45^{\circ}=1$, $\tan60^{\circ}=\sqrt3$, $\sin30^{\circ}=\frac12$, and $\cos45^{\circ}=\frac1{\sqrt2}$
$\frac{5\times(\frac12)^2+4(\frac2{\sqrt3})^2-1}{(\sqrt3)^2-(\frac12)^2-(\frac1{\sqrt2})^2}$
= $\frac{\frac54+\frac{16}{3}-1}{3-\frac14-\frac12}$
= $\frac{\frac{15+64-12}{12}}{\frac{12-1-2}{4}}$
= $\frac{67}{9\times3}$
= $\frac{67}{27}$
Hence, the correct answer is $\frac{67}{27}$.
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