Assuming that by tan^(-1)(4x) you mean arctan(4x) and not cot(4x).
We know the pythagorean identity sin^2(theta) + cos^2(theta) = 1, so if we divide both sides by #sin^2(theta) we'd have:
sin^2(theta)/sin^2(theta) + cos^2(theta)/sin^2(theta) = 1/sin^2(theta) rarr
1 + cot^2(theta) = csc^2(theta)
So, if we switch in theta for arctan(4x), we have
1 + 1/tan^2(arctan(4x)) = 1/sin^2(arctan(4x))
Knowing that tan(arctan(theta)) = theta
1 + 1/(4x)^2 = 1/sin^2(arctan(4x))
Taking the least common multiple so we can sum the two fractions on the right side:
(16x^2+ 1)/(16x^2) = 1/sin^2(arctan(4x))
Invert both sides
(16x^2)/(16x^2+1) = sin^2(arctan(4x))
Taking the square root
sqrt((16x^2)/(16x^2+1)) = sin(arctan(4x))
Simplifying
(4x)/sqrt(16x^2+1) = sin(arctan(4x))
You can rationalize it if you want
(4x*sqrt(16x^2+1))/(16x^2+1) = sin(arctan(4x))
And last but not least, gotta list out the values x can't be because of the liberities we took during our process. We can't have denominators being zero nor negatives in square roots, so we have:
(4x)^2 != 0
16x^2 +1 > 0
sin(arctan(4x)) != 0 rarr (4x)/sqrt(16x^2+1) != 0
From these we see that x != 0
