Question 519012
we know that tan 3x= sin3x/cos3x, let call this formula (a) 

       formulas to keep in mind before you start reading      
                  sin(a+b)= sin(a)cos(b)+ sin(b)cos(a),  let call it (1)
                  cos(a+b)= cos(a)cos(b)- sin(a)sin(b),  let call it (2)
                  cos(2x)= cos^2(x)- sin^2(x)
                  sin(2x)= 2cos(x)sin(x)
                  sin^2(x)+ cos^2(x)= 1
 
base on the (1), sin3x= sin(2x+x)
                      = sin(2x)cos(x)+ sin(x)cos(2x)
                      = 2 cos^2(x)sin(x)+ sin(x)[cos^2(x)-sin^2(x)]
                      = sin(x)[ 2cos^2(x)+ cos^2(x)- ( 1-cos^2(x) ]
                      = sin(x)[ 4cos^2(x)-1 ]
                      = [sqrt of (1-cos^2(x)][4cos^2(x)-1] 
                      = [ sqrt of (1-(3/5)^2) ][ 4*(3/5)^2 -1 ]
                      = 4/5 X 11/25
                      = 44/125

the (2) ,       cos3x = cos(2x+x)
                      = cos(2x)cos(x)- sin(x)sin(2x)
                      = [ cos^2(x)-sin^2(x)]cos(x) - 2sin^2(x)cos(x)
                      = cos(x)[ cos^2(x)-(1- cos^2(x)) - 2(1-cos^2(x)) ]
                      = cos(x)[ 4cos^2(x) - 3]
                      = 3/5[ 4*(3/5)^2 - 3]
                      = 3/5 X -39/25
                      = -117/125

so, tan 3x= 44/125:(-117/125)
          = - 44/117