`tan u = 3/5, 0

Given `tan(u)=3/5, 0<u<pi/2`

Angle u is in quadrant 1. A right triangle can be drawn is quadrant 1. Since 

`tan(u)=3/5` , the side opposite angle u is 3 and the side adjacent to angle u is 5. Using the pythagorean theorem the hypotenuse of the triangle is `sqrt34.`

`sin(2u)=2sin(u)cos(u)=2(3/sqrt34)(5/sqrt34)=30/34=15/17`

`cos(2u)=1-2sin^2(u)=1-2(3/sqrt34)^2=1-18/34=34/34-18/34=16/34=8/17`

`tan(2u)=(2tan(u))/(1-tan^2(u))=(2(3/5))/(1-(3/5)^2)=(6/5)/(1-9/25)=(6/5)/(25/25-9/25)=(6/5)/(16/25)=15/8`