If$$\ \frac{x-xtan^{2}15^\circ}{1+tan^{2}15^\circ}$$= sin $$60^\circ\ $$+ cos 30$$^\circ\ $$, then what is then what is the value of x?

$$tan15^\circ=\frac{\sqrt3-1}{\sqrt3+1}$$

Expression = $$\ \frac{x-xtan^{2}15^\circ}{1+tan^{2}15^\circ}$$= sin $$60^\circ\ $$+ cos 30$$^\circ\ $$

=> $$\ \frac{x(1-tan^{2}15^\circ)}{1+tan^{2}15^\circ}= \frac{\sqrt3}{2}+\frac{\sqrt3}{2}$$

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

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

=> $$x\times\frac{(3+1+2\sqrt3)-(3+1-2\sqrt3)}{(3+1+2\sqrt3)+(3+1-2\sqrt3)}=\sqrt3$$

=> $$x\times\frac{4\sqrt3}{8}=\sqrt3$$

=> $$x=\frac{8}{4}=2$$

=> Ans - (A)