If x, y are acute angles, where 0 < x + y < 90° and sin (3x - 40°) = cos(3y + 40°), then the value of tan(x + y) is equal to:

$\frac{1}{\sqrt{2}}$ $\frac{1}{\sqrt{3}}$ 1 $\frac{1}{3}$

$\frac{1}{\sqrt{3}}$

sin (3x - 40°) = cos(3y + 40°) Formula used :- sin A = cos B Then , A + B = 90º So, ( 3x - 40º  + 3y + 40º ) = 90º  x + y = 30º  Now,  tan( x + y ) = tan30º  =  \(\frac{1 }{ √3 }\)