Given $t = \tan \frac{\theta}{2}$, show $\sin \theta = \frac{2t}{1+t^2}$ Given $t = \tan \frac{\theta}{2}$, show $\sin \theta = \frac{2t}{1+t^2}$
There are a few ways to approach it, one of the way i encountered is that using the $\tan2\theta$ formula, we get $$\tan\theta = \frac{2t}{1-t^2}$$
By trigonometry, we know the ratio of the triangle, that is the opposite to adjacent is $2t : 1-t^2$, and hence it follows the hypotenuse is $\sqrt{(2t)^2+(1-t^2)^2} = \sqrt{(t^2+1)^2} = (t^2+1)$
My question is, can the above square rooted answer be $-(t^2+1)$ also? Why do we reject the negative answer? 

Because $$\frac{2t}{1+t^2}=\frac{2\tan\frac{\theta}{2}}{\frac{1}{\cos^2\frac{\theta}{2}}}=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=\sin\theta.$$
You can use your formula, but we need to write before:
$$\frac{1-t^2}{1+t^2}=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}=\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}=\cos\theta.$$
Id est,
$$\sin\theta=\tan\theta\cos\theta=\frac{2t}{1-t^2}\cdot\frac{1-t^2}{1+t^2}=\frac{2t}{1+t^2}.$$
