Calculus
posted by sh .
Find the value of tan2x, (pi/2)<x<pi, given secx=5/4.
So cosx=4/5? I've no clue what to do next.

According to the given domain, the angle must be in quadrant II,
We are given cosx =  4/5.
remember that cosine is adjacent/hypotenuse, so construct a right angles triangle with base of 4 and hypotenuse of 5, you should recognize the 345 rightangled triangle so the height (or opposite) is 3 and
sinx = 3/5
remember that tangent = sine/cosine
and tanx = (3/5)/(4/5) =  3/4
have you come across the identiy
tan 2x = 2tanx/(1  tan^2 x) ?
tan2x = 2(3/4)/(1  9/16)
= (3/2)/(7/16)
= 24/7 
Thanks!
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