Given $\tan\theta=7/9$ and $\cos\theta<0$, find $\tan(\theta/2)$.
Since the angle is lying on quadrant 3, I got
$$\cos\theta=-9/\sqrt{130}$$
and substitute it in
$$\tan(\theta/2)=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$$
However, there is a negative sign in the final answer
$$\tan(\theta/2)=-\sqrt{\frac{1+\frac{9}{\sqrt{130}}}{1-\frac{9}{\sqrt{130}}}}$$
I know when we square root a number, there will have a plus or minus sign outside the square root, but I don't know why it pick the negative one in this case.
Best Answer
Using $t$-formula
Let $t=\tan \dfrac{\theta}{2}$,
\begin{align*} \tan \theta &= \frac{2t}{1-t^2} \\ \frac{2t}{1-t^2} &= \frac{7}{9} \\ 7t^2+18t-7 &= 0 \\ t &= \frac{-9\pm \sqrt{130}}{7} \\ \cos \theta &= \frac{1-t^2}{1+t^2} \\ \frac{1-t^2}{1+t^2} &< 0 \\ t^2 &> 1 \end{align*}
Rejecting $t = \dfrac{-9+\sqrt{130}}{7}$, we have