tan of a Half Angle

Divide the half-angle sine by the half-angle cosine and tidy up, and the half-angle tangent needs no square root at all:

$$\tan\left(\frac{x}{2}\right) = \frac{1-\cos\left(x\right)}{\sin\left(x\right)}$$

$$\tan\left(\frac{x}{2}\right) = \frac{\sin\left(x\right)}{1+\cos\left(x\right)}$$

Both hold wherever both sides are defined. With x = 2A they turn a double angle back into A: $\frac{\sin\left(2A\right)}{1+\cos\left(2A\right)}$ is $\tan\left(A\right)$.

At 90°: $\frac{1-\cos\left(90^{\circ}\right)}{\sin\left(90^{\circ}\right)} = 1$, and tan 45° is indeed 1.