Question 1015107
<pre>
Are you sure you didn't mean tan(2x+y)???


{{{y = 2 ("180°"n-x+"60°")}}}

Use the identity:

If sin(A) = cos(B), then A = 90°-B+360°n

sin(2x-20°)= cos(y-10°)

2x-20°= 90°-(y-10°)+360°n

2x-20°= 90°+y+10°+360°n

2x-20°= 100°+y+360°n

2x = 120°+y+360°n

2x-120°-360°n = y

x+y = x+(2x-120°-360°n)

x+y = 3x-120°-360°n 

tan(x+y) = tan(3x-120°-360°n)

tan(x+y) = tan(3x-120°) = {{{( tan(3x)-tan("120°") )/( 1+tan(3x)tan("120°") )}}}  


 = {{{( tan(3x)-(-sqrt(3)) )/( 1+tan(3x)(-sqrt(3)) )}}}

 = {{{( tan(3x)+sqrt(3) )/( 1-tan(3x)(sqrt(3)) )}}}

tan(2x+y) works out much nicer.

Edwin</pre>