<s>[INST] 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.
 [/INST] If $\alpha=\cos^{-1}\left(\sqrt{\frac{2+\sqrt3}4}\right)$, then $4\cos^2\alpha=2+\sqrt3$.
This means $\sqrt3=4\cos^2\alpha-2=2(2\cos^2\alpha-1)=2\cos2\alpha$.
$2\alpha=\cos^{-1}\frac{\sqrt3}2=\frac{\pi}6$. So $\alpha=\frac\pi{12}$.
So...

 $T=\sin^{-1}\cot(\frac\pi2)=\sin^{-1}0=0$.

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