How do you evaluate $cot[cos^-1(sqrt5/6)]$ ?

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Answer 1 · 2016-07-17T04:09:35

0.402

Use calculator.
$sqrt5/6 = 0.373$
cos x = 0.373 --> arc $x = 68^@12$
sin x = sin 68.12 = 0.927
$cot x = (cos x)/(sin x) = 0.373/0.927 = 0.402$

How do you evaluate cot[cos^-1(sqrt5/6)]cot[cos^-1(sqrt5/6)]?