Question

Prove the following identity. [sin(x+y)+sin(x-y)]/[cos(x+y)+cos(x-y)] = tan(x)

Answer 1

$\frac{sin(x + y) + sin(x - y)}{cos(x + y) + cos(x - y)} = tan(x)$
$\frac{(sin(x)cos(y) + cos(x)sin(y)) + (sin(x)cos(y) - cos(x)sin(y))}{(cos(x)cos(y) - sin(x)sin(y)) + (cos(x)cos(y) + sin(x)sin(y))} = tan(x)$
$\frac{(sin(x)cos(y) + sin(x)cos(y)) + (cos(x)sin(y) - cos(x)sin(y))}{(cos(x)cos(y) + cos(x)cos(y)) + (-sin(x)sin(y) + (sin(x)sin(y))} = tan(x)$
$\frac{2sin(x)cos(y)}{2cos(x)cos(y)} = tan(x)$
$\frac{sin(x)}{cos(x)} = tan(x)$