`tan A = 1/7`
`:. sec A = sqrt(1+tan^2A) = sqrt(1+1/49) = (5sqrt2)/7`
`:. cosA = 7/(5sqrt2)`
`:. sin A = tanAcosA = 1/(5sqrt2)`
Now, `tanB = 1/3`
`:. sec B = sqrt(1+tan^2B) = sqrt(1+1/9) = (sqrt10)/3`
`:. cosB = 3/(sqrt10)`
`:. sin B = tanBcosB = 1/sqrt10`
Now, `L.H.S. = cos2A = 2cos^2A -1 = 2((7/5sqrt2)^2 - 1 = 98/50-1 = 48/50 = 24/25`
`R.H.S. = sin4B = 2sin2Bcos2B = 2(2sinBcosB)(2cos^2B-1)`
`=2(2(1/sqrt10)(3/sqrt10))(2(9/10)-1)`
`=2(6/10)(4/5) = 24/25`
`:. cos2A = sin4B = 24/25`
