`0 < A ,B < pi/4 ,cos (A+B) =4/5, sin (A-B) =5/13 => tan 2A=`

Explanation

`0 lt A, B lt pi/4`
`rArr -pi/4 lt A-B lt pi/4` and `0 lt A + B lt pi/2`
Now `sec(A+B)=1/(cos(A+B))=5/4`
`rArr tan^(2)(A+B)=sec^(2)(A+B)-1 =(5/4)^(2)-1=9/16`
`rArr tan(A+B)=3/4`
(tan is positive in first quadrant).
`cos^(2)(A-B)=1-sin^(2)(A-B)`
`=1-(5/13)^(2)=144/169`
`therefore cos(A-B)=12/13`,
In`-pi/4, pi/4` cos is positive.
and `tan(A-B)=(sin(A-B))/(cos(A-B))=(5/13)/(12/13)=5/12`
Now `tan2A=tan{(A+B)+(A-B)}`
`=(tan(A+B)+tan(A-B))/(1-tan(A+B)tan(A-B))`
`=(3/4+5/12)/(1-3/4.5/12)=56/33` Ans.