Find the value of ${{{\cos }^{ - 1}}\left( {\sqrt {\frac{{2 + \sqrt 3 }}{4}} } \right)}$ I am trying to solve:
${\sin ^{ - 1}}\cot \left( {{{\cos }^{ - 1}}\left( {\sqrt {\frac{{2 + \sqrt 3 }}{4}} } \right) + {{\cos }^{ - 1}}\left( {\frac{{\sqrt {12} }}{4}} \right) + \csc{^{ - 1}}\left( {\sqrt 2 } \right)} \right)$
My solution is as follow:
$T = {\sin ^{ - 1}}\cot \left( {{{\cos }^{ - 1}}\left( {\sqrt {\frac{{2 + \sqrt 3 }}{4}} } \right) + {{\cos }^{ - 1}}\left( {\frac{{\sqrt {12} }}{4}} \right) + \csc{^{ - 1}}\left( {\sqrt 2 } \right)} \right) $
Since:
$\csc{^{ - 1}}\left( {\sqrt 2 } \right) = {\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{\pi }{4};{\cos ^{ - 1}}\left( {\frac{{\sqrt {12} }}{4}} \right) = {\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = \frac{\pi }{6}$
Then:
$T = {\sin ^{ - 1}}\cot \left( {{{\cos }^{ - 1}}\left( {\sqrt {\frac{{2 + \sqrt 3 }}{4}} } \right) + \frac{\pi }{4} + \frac{\pi }{6}} \right)$
I am not able to proceed further.

\begin{align}
\theta &= \cos^{-1}\sqrt{\frac{2+\sqrt{3}}{4}}\\
\cos \theta &=  \sqrt{\frac{2+\sqrt{3}}{4}}\\
\cos^2 \theta &= \frac{2+\sqrt{3}}{4}\\
\frac 12 + \frac 12 \cos 2\theta &= \frac{2+\sqrt{3}}{4}\\[0.5em]
2 + 2 \cos 2\theta &= 2+ \sqrt 3\\
\cos 2\theta &= \frac{\sqrt 3}{2}\\
2\theta &= \frac{\pi}{6}\\[1em]
\theta &= \frac{\pi}{12}
\end{align}
