If sin (x + y) = cos [3(x + y)], then the value of tan[2(x + y)] is :

Given,

sin (x + y) = cos [3(x + y)]
Using: cos θ = sin (90° - θ)
sin (x + y) = sin [90° - 3(x + y)]
sin [90° - 3(x + y)] - sin (x + y) = 0
sinC−sinD =2sin[(C−D)/2] cos[(C+D)/2]
=2sin$$\frac{(90-3(x+y)-(x+y))}{2} cos \frac{90-3(x+y)+(x+y)}{2}$$=0
=2sin(45-2(x+y)) cos (45-(x+y))=0
∴ sin 45° - 2(x + y)} = 0
45° - 2(x + y) = 0
2(x + y) = 45°
OR
Cos{45°- (x + y)} = 0
45°- (x + y)} = 90°
x + y = - 45°
2(x + y) = - 90°
Putting 2(x + y) = 45°
tan 2(x + y) = tan 45° = 1
Again, Putting 2(x + y) = - 90°, we will not get any answer among given options
Option B is the correct answer.