Cool Trigonometry

Let $t = \tan{\alpha} = \dfrac{\sqrt{3}+1}{\sqrt{3}-1} = \dfrac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)} = \dfrac{4+2\sqrt{3}}{2} = 2 + \sqrt{3}$

$\begin{aligned} \cos{2\alpha} + (2+\sqrt{3})\sin{2\alpha} & = \frac{1-t^2}{1+t^2} + t \dot{} \frac{2t}{1+t^2} = \frac{1+t^2}{1+t^2} = \boxed{1} \end{aligned}$

$tg(\alpha)=2+\sqrt{3}$ . Therefore $cos (2\alpha) + (2+\sqrt{3})sin (2\alpha)$ $= cos^{2} (\alpha) - sin^{2} (\alpha)+\frac{sin (\alpha)}{cos (\alpha)}\cdot 2sen (\alpha)cos (\alpha)$ $=cos^{2}+sin^{2}=1$

For my convinience I would be taking 'alpha' as 'x',

After rationalising the given term,

tan x = 2 + (3)^0.5

sin x =[2 + (3)^0.5]cos x

Now the expression whose value we need to find,

cos 2x + (2 + (3)^0.5)sin 2x

cos^2 x - sin^2 +(2)[2 + (3)^0.5]cos x . sin x

cos^2 x - sin^2 x + 2 sin x . sin x

cos^2 x - sin^2 + 2sin^2 x

cos^2 x + sin^2 x = 1

Based on the given information $\tan { \alpha }$ , create a triangle of sides length $\sqrt { 3 } +1$ and $\sqrt { 3 } -1$ . Thus, the hypotenuse of the triangle, after calculating with Pythagoras theorem, is $\sqrt { 8 }$ . Now the values of $\sin{ \alpha }, \cos{ \alpha }$ and $\tan{ \alpha }$ can be calculated simply by referring to this triangle (which you have hopefully ingrained in your head by now).

$\cos { 2\alpha }$ can be calculated with:

$\cos{ 2\alpha }=\cos ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \alpha } =\frac { 4-2\sqrt { 3 } }{ 8 } -\frac { 4+2\sqrt { 3 } }{ 8 } =-\frac { \sqrt { 3 } }{ 2 }$

$\sin { 2\alpha }$ can be calculated with:

$\sin { 2\alpha } =2\sin { \alpha } \cos { \alpha } =2(\frac { \sqrt { 3 } +1 }{ \sqrt { 8 } } )(\frac { \sqrt { 3 } -1 }{ \sqrt { 8 } } )=\frac { 1 }{ 2 }$

After substituting the values into the expression and simplifying it, you are left with $\boxed{ 1 }$ .