How do you evaluate $tan (A r c cos (\frac{\sqrt{2}}{2}))$ $tan (Arc cos (sqrt2/2) )$ ?

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How do you evaluate $tan (A r c cos (\frac{\sqrt{2}}{2}))$ $tan (Arc cos (sqrt2/2) )$ ?

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Answer 1 · 2016-06-08T12:44:52

$\pm 1$ $+-1$ . It is 1, if $a = a r c cos (\frac{\sqrt{2}}{2})$ $a = arc cos (sqrt 2/2)$ is restricted to be in the 1st quadrant...

Explanation:

Let $a = a r c cos (\frac{\sqrt{2}}{2}) = a r c cos (\frac{1}{\sqrt{2}})$ $a=arc cos (sqrt 2/2)=arc cos (1/sqrt 2)$ . As cos (+-a)= cos a, a is in the 1st quadrant or in the 4th. Accordingly, $sin a = +-1/sqrt 2 and, therefore, tan a =+-1$ .

if $a = arc cos (sqrt 2/2)$ is restricted to be in the 1st quadrant. the given expression tan a = 1...

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How do you evaluate $tan (A r c cos (\frac{\sqrt{2}}{2}))$ $tan (Arc cos (sqrt2/2) )$ ?

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How do you evaluate $tan (A r c cos (\frac{\sqrt{2}}{2}))$ $tan (Arc cos (sqrt2/2) )$ ?