#sinx = "opposite side"/"hypotenuse"#
Let #k# be the common multiple,
Hence, one of the sides (opposite side) will be #1k# and the hypotenuse will be, #sqrt3k#
So the other (adjacent side) would be #sqrt((sqrt3k)^2- (k)^2# #["applying Pythagoras theorem"]#
#=> sqrt(2k^2) = sqrt2 k#
#color(white)(wwww#
Now, #tanx = "opposite side"/"adjacent side" = 1/sqrt2#
#tan2x= (2tanx)/(1-tan^2x)#
#tan2x= (2(1/sqrt2))/(1-(1/2))#
#tan2x= (sqrt2)/(1/2) = 2sqrt2#
Alternate way,
#tan2x=(sin2x)/(cos2x) = (2sinxcosx)/(1-2sin^2x)#
#=> (2(1/sqrt3)(sqrt2/sqrt3))/(1-2(1/3))#
#=> ((2sqrt2)/3)/((3-2)/3) = 2sqrt2#
