Verified answer

I'll use arccos(v) instead of cos^-1v:

tan(arccos(v))

= sin(arccos(v)) / cos(arccos(v))

by definition of arccos, cos(arcos(v)) = v

= sin(arccos(v)) / v

for the sin, using the identity: (sin(x))^2 + (cos(x))^2 = 1, we can rewrite sin(arccos(v)) in a different form:

(sin(arccos(v)))^2 + (cos(arccos(v)))^2 = 1

(sin(arccos(v)))^2 = 1 - (cos(arccos(v)))^2

sin(arccos(v)) = sqrt(1 - (cos(arccos(v)))^2)

sin arccos(v) gives an angle between 0 <= theta <= 180, so sin(arccos(v)) must be positive since sin of any angle between 0 and 180 is positive. So the right side sqrt(1 - (cos(arccos(v)))^2) must be positive. So we dont have to the plus-minus sign in front of the square root since we know its positive.

Now replace sin(arccos(v)) with sqrt(1 - (cos(arccos(v)))^2) :

= sqrt(1 - (cos(arccos(v)))^2) / v

= sqrt(1 - v^2) / v

put the v in the denominator into the square root by squaring the v.

= sqrt((1 - v^2) / v^2)

= sqrt(1/v^2 - 1)