Question : If $\frac{x-x\tan^{2}30^{\circ}}{1+\tan^{2}30^{\circ}}=\sin^{2}30^{\circ}+4\cot^{2}45^{\circ}-\sec^{2}60^{\circ}$, then value of $x$ is:
Option 1: $\frac{1}{4}$
Option 2: $\frac{1}{5}$
Option 3: $\frac{1}{2}$
Option 4: $\frac{1}{\sqrt3}$
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Correct Answer: $\frac{1}{2}$
Solution :
Given the equation,
$\frac{x-x\tan^{2}30^{\circ}}{1+\tan^{2}30^{\circ}}=\sin^{2}30^{\circ}+4\cot^{2}45^{\circ}-\sec^{2}60^{\circ}$
We know that $\tan30^{\circ} = \frac{1}{\sqrt{3}}$, $\sin30^{\circ} = \frac{1}{2}$, $\cot45^{\circ} = 1$, and $\sec60^{\circ} = 2$.
$⇒\frac{x-x(\frac{1}{3})}{1+( \frac{1}{3})}=\left(\frac{1}{2}\right)^2+4(1)^2-(2)^2$
$⇒\frac{\frac{2x}{3}}{\frac{4}{3}}=\frac{1}{4}+4-4$
$⇒\frac{1}{2}x=\frac{1}{4}$
$⇒x=\frac{1}{2}$
Hence, the correct answer is $\frac{1}{2}$.
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