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Question 140
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:
Solution
Expression : $$\frac{x-x\ tan^{2}\ 30^\circ}{1+tan^{2}\ 30^\circ}=sin^{2}\ 30^\circ+4\ cot^{2}\ 45^\circ-sec^{2}\ 60^\circ$$
=>Â $$\frac{x-x(\frac{1}{\sqrt3})^2}{1+(\frac{1}{\sqrt3})^2}=(\frac{1}{2})^2+4(1)^2-(2)^2$$
=> $$\frac{\frac{2x}{3}}{\frac{4}{3}}=\frac{1}{4}+4-4$$
=> $$\frac{x}{2}=\frac{1}{4}$$
=> $$x=\frac{2}{4}=\frac{1}{2}$$
=> Ans - (D)
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