2014/071) prove $\cot \ 7\frac{1}{2}^\circ = \sqrt{2} + \sqrt{3}+\sqrt{4} + \sqrt{6}$

we have

$\cos\ 15^\circ = \cos(60- 45)^\circ =  \cos\ 60^\circ \cos\ 45^\circ +  \sin \ 60^\circ \sin\ 45^\circ$
= $\dfrac{1}{2}\dfrac{1}{\sqrt{2}} + \dfrac{\sqrt{3}}{2}\dfrac{1}{\sqrt{2}}$
= $\dfrac{1}{4}(\sqrt{2} + \sqrt{6})$
also
$\sin \ 15^\circ = \sin (60- 45)^\circ =  \sin \ 60^\circ \cos\ 45^\circ -  \cos  \ 60^\circ \sin\  45^\circ$
= $\dfrac{\sqrt{3}}{2}\dfrac{1}{\sqrt{2}} - \dfrac{1}{2}\dfrac{1}{\sqrt{2}}$
= $\dfrac{1}{4}(\sqrt{6} - \sqrt{2})$

now $\cot\ x= \dfrac{\cos\ x }{\sin \ x} = \dfrac{2\cos^2\ x }{2\sin \ x\cos \ x}$
= $\dfrac{1+\cos 2x}{\sin 2x}$

  Hence $\cot \ 7\frac{1}{2}^\circ = \dfrac{1+\cos\ 15^\circ}{\sin\ 15^\circ}$
= $\dfrac{1+\dfrac{1}{4}(\sqrt{2} + \sqrt{6})}{\dfrac{1}{4}(\sqrt{6} - \sqrt{2})}$
=  $\dfrac{4+(\sqrt{2} + \sqrt{6})}{(\sqrt{6} - \sqrt{2})}$
= $\dfrac{4+(\sqrt{6} + \sqrt{2})}{(\sqrt{6} - \sqrt{2})}$
= $\dfrac{4(\sqrt{6} + \sqrt{2})+(\sqrt{6} + \sqrt{2})^2}{(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2})}$
=  $\dfrac{4(\sqrt{6} + \sqrt{2})+(\sqrt{6} + \sqrt{2})^2}{4}$
=  $(\sqrt{6} + \sqrt{2})+\dfrac{(\sqrt{6} + \sqrt{2})^2}{4}$
=  $(\sqrt{6} + \sqrt{2})+\dfrac{6 + 2 + 4 \sqrt{3}}{4}$
=  $(\sqrt{6} + \sqrt{2})+\dfrac{8 + 4 \sqrt{3}}{4}$
=  $(\sqrt{6} + \sqrt{2})+  2 +  \sqrt{3}$
=  $\sqrt{2} + \sqrt{3})+  2 +  \sqrt{6}$
=  $\sqrt{2} + \sqrt{3})+\sqrt{4} + \sqrt{6}$