Find The Value Of Tan 15 Degree Class 11

$\\Solution:\\ \tan (A+ B)=\frac{\tan A+\tan B}{1-\tan A.\tan B}\\\\\\ \Rightarrow \tan(60^0+45^0)=\frac{\tan 60^0+\tan 45^0}{1-\tan 60^0.\tan 45^0}\\\\\\ \Rightarrow \tan(105^0)=\frac{\sqrt3+1}{1-\sqrt3\times1}\\\\\\ \Rightarrow \tan(105^0)=\frac{\sqrt3+1}{1-\sqrt3}\times\frac{1+\sqrt3}{1+\sqrt3}\\\\\\ \Rightarrow \tan(105^0)=\frac{(\sqrt3+1)^2}{1^2-\sqrt3^2}\ \ \ \ \ \ \ [\because (a+b)(a-b)=a^2-b^2]\\\\\\ \Rightarrow \tan(105^0)=\frac{3+1+2\sqrt3}{1-3}\ \ \ \ \ \ \ [\because (a+b)^2=a^2+b^2+2ab]\\\\\\ \Rightarrow \tan(105^0)=\frac{4+2\sqrt3}{-2}\\\\\\ \Rightarrow \tan(105^0)=\frac{2(2+\sqrt3)}{-2}\\\\\\ \therefore \tan(105^0)=-(2+\sqrt3)$