What does $sec (arccos (2)) + 3 cot (arctan (- 1))$ $sec(arccos(2))+3cot(arctan(-1))$ equal?

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Answer 1 · 2016-04-12T12:32:30

As absolute value of cos x = |cos x| $\leq 1$ $<=1$ , there is no x for which cosine is 2.

If x = arc cos (2), cos x = 2 > 1.
So, arc cos(2) is unreal..

The second term = $3 cot (a r c cot (\frac{1}{- 1}) = 3 cot (a r c cot (- 1)) = 3 (- 1) = - 3$ $3cot(arc cot(1/(-1))=3cot(arc cot(-1))=3(-1)=-3$

Successive operations by 'cot' over 'arc cot' amounts to multiplication by 1..

What does sec(arccos(2))+3cot(arctan(−1))sec(arccos(2))+3cot(arctan(-1)) equal?