This is an angle in a Pythagorean triangle (5,12,13) In Quadrant I where all sides are positive. SinA=5/13 and cosA=12/13 so tanA=5/12.
There are two ways to find tan2A.
- Use sin2A/cos2A=2sinAcosA/(2cos^2(A)-1)=2*5/13*12/13÷(2*144/169-1)=120/(288-169)=120/119.
- Use tan2A=2tanA/(1-tan^2(A))=5/6÷(1-25/144)=5/(6-25/24)=120/(144-25)=120/119.