# If $\frac{1+\sin{\theta}}{\cos{\theta}} = n$, find $\tan{\frac{1}{2}\theta}$.

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.

• (Co)tangent half-angle formula and cofunction identities.
– anon
Commented Nov 26, 2022 at 5:14

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{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}

Cross-multiplying and simplifying then gives

\begin{aligned} n - n\tan(\alpha) & = 1 + \tan(\alpha) \\ (-n - 1)\tan(\alpha) & = 1 - n \\ \tan(\alpha) & = \frac{n-1}{n+1} \end{aligned}

• Thanks a lot! I didn’t think to make $\alpha = \frac{\theta}{2}$. Commented Nov 26, 2022 at 5:53

Split LHS into two terms. We get $$n-\tan \theta =\sec \theta$$. Squaring we get $$n^{2}+\tan^{2}\theta-2n\tan \theta=1+\tan^{2}\theta$$. Can you continue?

• I'll try for now. Thanks. Commented Nov 26, 2022 at 5:13
• I got $\tan{\theta}=\dfrac{n^2-1}{2n}$. How do I advance from here? Commented Nov 26, 2022 at 5:23
• @nich $\tan \theta=\frac {2\tan (\frac {\theta}2)} {1-\tan^{2}(\frac {\theta}2 )}$ Commented Nov 26, 2022 at 6:11

If we also have the "tangent half-angle" identity $$\ \tan \frac{\theta}{2} \ = \ \frac{1 \ - \ \cos \theta}{\sin \theta} \ \ , \$$ we can approach the problem in this way. From $$\ n \ = \ \frac{1 \ + \ \sin \theta}{\cos \theta} \ = \ \sec \theta + \tan \theta \ \Rightarrow \ \tan \theta \ = \ n - \sec \theta \ \ , \$$ construct a right triangle with this tangent "value". (We will produce an expression for the hypothenuse $$\ H \$$ shortly.)

It is clear that $$\ \sec \theta \ = \ H \ \ . \$$ We then have

$$\ \tan \frac{\theta}{2} \ \ = \ \ \csc \theta \ - \ \cot \theta \ \ = \ \ \frac{H \ - \ 1}{n \ - \ H} \ \ .$$

From the "Pythagorean theorem", we find $$H^2 \ - \ 1 \ \ = \ \ (n \ - \ H)^2 \ \ = \ \ n^2 \ - \ 2nH \ + \ H^2 \ \ \Rightarrow \ \ nH \ \ = \ \ \frac{n^2 \ + \ 1}{2} \ \ .$$

Consequently, $$\ \tan \frac{\theta}{2} \ \ = \ \ \frac{nH \ - \ n}{n^2 \ - \ nH} \ \ = \ \ \frac{\left(\frac{n^2 \ + \ 1}{2} \right) \ - \ n}{n^2 \ - \ \left(\frac{n^2 \ + \ 1}{2} \right)} \ \ = \ \ \frac{ n^2 \ + \ 1 \ - \ 2n}{2n^2 \ - \ ( n^2 \ + \ 1 )}$$ $$= \ \ \frac{ n^2 \ - \ 2n \ + \ 1 }{ n^2 \ - \ 1 } \ \ = \ \ \frac{ (n \ - \ 1)^2 }{ (n \ + \ 1) · (n \ - \ 1) } \ \ = \ \ \frac{ n \ - \ 1 }{ n \ + \ 1 } \ \ .$$

[This contains elements appearing in the other answers, arrived at in a different way. This is perhaps inevitable since all of them hinge on the "Pythagorean identity" $$\ \tan^2 \theta \ + \ 1 \ = \ \sec^2 \theta \$$ in some manner.]