The question is:
If $\cfrac{1+\sin{\theta}}{\cos{\theta}} = n$, find $\tan{\frac{1}{2}\theta}$.
I'm quite confused on how to solve this problem. What trigonometric identities do I use? I tried squaring them and ended with $n^2=\dfrac{1+\sin{\theta}}{1-\sin{\theta}}$ but I suppose that doesn't really help the problem. Any suggestions would be appreciated.
Best Answer
Let $\alpha = \frac{\theta}{2} \; \to \; \theta = 2\alpha$, and use the double-angle formulae to convert from $\sin(2\alpha)$ and $\cos(2\alpha)$ to functions of $\tan(\alpha)$, to get
$$\begin{equation}\begin{aligned} n & = \frac{1 + \sin(2\alpha)}{\cos(2\alpha)} \\ & = \frac{1 + \frac{2\tan(\alpha)}{1+\tan^2(\alpha)}}{\frac{1-\tan^2(\alpha)}{1+\tan^2(\alpha)}} \\ & = \frac{(1+\tan^2(\alpha))+2\tan(\alpha)}{1-\tan^2(\alpha)} \\ & = \frac{(1+\tan(\alpha))(1+\tan(\alpha))}{(1-\tan(\alpha))(1+\tan(\alpha))} \\ & = \frac{1+\tan(\alpha)}{1-\tan(\alpha)} \end{aligned}\end{equation}$$
Cross-multiplying and simplifying then gives
$$\begin{equation}\begin{aligned} n - n\tan(\alpha) & = 1 + \tan(\alpha) \\ (-n - 1)\tan(\alpha) & = 1 - n \\ \tan(\alpha) & = \frac{n-1}{n+1} \end{aligned}\end{equation}$$