Very quickly: since tan(30°) = 1/√3,
(1 + tan(30°)) / (1 - tan(30°)) = (1 + 1/√3) / (1 - 1/√3)
… = (√3 + 1) / (√3 - 1)
… = (√3 + 1)² / ((√3)² - 1²)
… = ((√3)² + 2√3 + 1²) / (3 - 1)
… = (3 + 2√3 + 1) / 2
… = (4 + 2√3) / 2
… = 2 + √3
On the other side, we have sin(30°) = 1/2 and sin(60°) = √3/2, so
(1 + sin(60°)) / (1 - sin(30°)) = (1 + √3/2) / (1 - 1/2)
… = (1 + √3/2) / (1/2)
… = 2 (1 + √3/2)
… = 2 + √3
and we're done, both sides are equal.
We can also prove this algebraically:
Force the denominator on the left side into a difference of squares:
(1 + tan(30°)) / (1 - tan(30°)) × (1 + tan(30°)) / (1 + tan(30°))
… = (1 + tan(30°))² / (1 - tan²(30°))
Expand the numerator:
… = (1 + 2 tan(30°) + tan²(30°)) / (1 - tan²(30°))
Recall that 1 + tan²(x) = sec²(x) :
… = (sec²(30°) + 2 tan(30°)) / (1 - tan²(30°))
Since sec(x) = 1/cos(x) and tan(x) = sin(x)/cos(x), multiply through every term by cos²(30°) :
… = (sec²(30°) + 2 tan(30°)) / (1 - tan²(30°)) × cos²(30°) / cos²(30°)
… = (1 + 2 sin(30°) cos(30°)) / (cos²(30°) - sin²(30°))
Recall that sin(2x) = 2 sin(x) cos(x) and cos(2x) = cos²(x) - sin²(x) :
… = (1 + sin(2×30°)) / cos(2×30°)
… = (1 + sin(60°)) / cos(60°)
Finally, recall that cos(90° - x) = sin(x), so cos(60°) = sin(30°) = 1/2, and we can write 1/2 = 1 - 1/2, so
… = (1 + sin(60°)) / (1 - sin(30°))
as required.