How do you find the value of $tan [a r c sec (\frac{\sqrt{5}}{2})]$ $tan [arcsec((sqrt 5)/2)]$ ?

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How do you find the value of $tan [a r c sec (\frac{\sqrt{5}}{2})]$ $tan [arcsec((sqrt 5)/2)]$ ?

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Answer 1 · 2016-06-21T13:27:01

$\frac{1}{2}$ $1/2$

Explanation:

might be clearer if it was broken down a bit

we want to find $tan θ$ $tan theta$ where $θ = a r c sec (\frac{\sqrt{5}}{2})$ $theta = arcsec (sqrt5 /2)$ so $sec θ = (\frac{\sqrt{5}}{2})$ $sec theta = (sqrt5 /2)$ so $cos θ = \frac{2}{\sqrt{5}} = \frac{a d j a c e n t}{h y p o t e N e u s e}$ $cos theta = 2/(sqrt5 ) = (adjacent) / (hypoteNeuse)$

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

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How do you find the value of $tan [a r c sec (\frac{\sqrt{5}}{2})]$ $tan [arcsec((sqrt 5)/2)]$ ?

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How do you find the value of $tan [a r c sec (\frac{\sqrt{5}}{2})]$ $tan [arcsec((sqrt 5)/2)]$ ?