How do you find the value of $tan [arcsec((sqrt 5)/2)]$ ?

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Answer 1 · 2016-06-21T13:27:01

$1/2$

might be clearer if it was broken down a bit

we want to find $tan theta$ where $theta = arcsec (sqrt5 /2)$ so $sec theta = (sqrt5 /2)$ so $cos theta = 2/(sqrt5 ) = (adjacent) / (hypoteNeuse)$

the opposite side is therefore $sqrt{5 - 2^2}= 1$ so $tan theta = 1/2$

How do you find the value of tan [arcsec((sqrt 5)/2)] tan [arcsec((sqrt 5)/2)]?