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 : $$\frac{x-x\ tan^{2}\ 30^\circ}{1+tan^{2}\ 30^\circ}=sin^{2}\ 30^\circ+4\ cot^{2}\ 45^\circ-sec^{2}\ 60^\circ$$

=> $$\frac{x-x(\frac{1}{\sqrt3})^2}{1+(\frac{1}{\sqrt3})^2}=(\frac{1}{2})^2+4(1)^2-(2)^2$$

=> $$\frac{\frac{2x}{3}}{\frac{4}{3}}=\frac{1}{4}+4-4$$

=> $$\frac{x}{2}=\frac{1}{4}$$

=> $$x=\frac{2}{4}=\frac{1}{2}$$

=> Ans - (D)

Create a FREE account and get: