<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] Hint:
$$y=\dfrac{2+\sqrt3}4=\dfrac{(\sqrt3+1)^2}8$$
$$\sqrt y=\dfrac{\sqrt3+1}{2\sqrt2}=\cos\dfrac\pi6\cos\dfrac\pi4+\sin\dfrac\pi6\sin\dfrac\pi4=\cos\left(\dfrac\pi4-\dfrac\pi6\right)$$
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