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 3-4-5 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)