trig
posted by Amandeep .
tanx = 5/12 and sinx<0
Find sin2x and cos2x
How to get the answers of sin2x = 120/169 and cos 2x = 119/169?

tanx = 5/12
r^2 = x^2 + y^2
= 12^2 + 5^2
r = 13
tanx >0 and sinx<0, so x must be in quad. III and
and sinx = 5/13 , cosx = 12/13
sin 2x = 2sinxcosx = 2(5/13)(12/13)
= 120/169
cos 2x = cos^2 x  sin^2 x
= 144/16925/169
=119/169
do the others the same way, or the way
drwls showed you in the other post. 
tan x = 13 cot x

tan x = 13 cot x
tan x = 13/tan x
tan^2 x = 13
tan x = ±√13
x = 74.5º,105.5º, 254.5º, 285.5º