Trig Identity Proof $\frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = 2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)$

Question

I've been working on this for like half an hour now and don't seem to be getting anywhere. I've tried using the double angle identity to write the LHS with $\theta/2$ and have tried expanding the RHS as a sum. The main issue I'm having is working out how to get a $\pi$ on the LHS or removing it from the right to equate the two sides. $$\frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = 2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)$$

The procedure you have suggested should work. Please show what you have done and explain clearly where you are stuck. — David, Jun 30 '17 at 01:17
To help show your work better, here's a starting point into the instructions for how to format math formulas on this site: https://math.stackexchange.com/help/notation — David K, Jun 30 '17 at 01:30
Have you tried to show the two quotients on the left are equal, and used the addition theorem for tan on the right? — Somos, Jun 30 '17 at 01:56
Yeah I got up to that part but don't know how to get rid of the pi/4 — juA, Jun 30 '17 at 01:57
Did you use the fact that $\tan(\pi/4) = 1$? Also, could you show us what you have done? — N. F. Taussig, Jun 30 '17 at 02:00
Could be useful to know that $\tan(\theta/2)=\frac {\sin\theta}{1+\cos\theta}$ — Mr. F, Jun 30 '17 at 02:13
Lol I'm an idiot. How did i not realise that tan45 was 1 xd. The question is easy now — juA, Jun 30 '17 at 02:14

score 3 · Accepted Answer · edited Jun 30 '17 at 03:17

3

hint: \begin{align} & \cos\theta = \cos^2(\theta /2) - \sin^2(\theta/2)= (\cos (\theta/2) +\sin (\theta/2))(\cos (\theta/2) - \sin (\theta/2)), \\[10pt] & 1 \pm\sin \theta = (\cos (\theta/2) \pm \sin(\theta/2))^2. \end{align}

edited Jun 30 '17 at 03:17

Michael Hardy

1

answered Jun 30 '17 at 01:18

DeepSea

77,651

Ok, I understand the first formula. How did you get the second one (after the comma)? – juA Jun 30 '17 at 01:31
To juA :.....$(\cos A/2 \pm \sin A/2)^2=$ $(\cos^2 A/2+\sin^2 A/2)\pm 2\cos A/2 \sin A/2=$ $1\pm 2\cos A/2 \sin A/2=$ $1\pm \sin A.$ – DanielWainfleet Jun 30 '17 at 02:59
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Here's the derivation of the part after the comma in some detail: $$ \begin{align} & \left( \cos\left( \frac\theta2 \right) \pm \sin\left( \frac\theta2\right) \right)^2 \ \ = {} & \cos^2 \left(\frac\theta2\right) \pm 2\cos\left( \frac\theta2\right)\sin \left( \frac\theta2\right) + \sin^2 \left( \frac\theta2 \right) \ \ = {} & 1 \pm 2 \cos \left( \frac\theta2 \right) \sin \left( \frac\theta2\right) \quad \text{since } \sin^2\alpha+\cos^2\alpha = 1\ \ = {} & 1 \pm \sin\theta \qquad \text{by the double-angle formula for the sine function.} \end{align} $$ $${}$$ – Michael Hardy Jun 30 '17 at 03:21
Nice. I actually did exactly that in my proof, I just didn't treat it as an identity and go straight to it – juA Jun 30 '17 at 06:47

score 2 · Answer 2 · answered Jun 30 '17 at 06:42

2

Let $\dfrac\theta2+\dfrac\pi4=y\implies\theta=2y-\dfrac\pi2$

$$\sin\theta=\cdots=-\cos2y,\cos\theta=\cdots=\sin2y$$

$$\dfrac{1+\sin\theta}{\cos\theta}=\dfrac{1-\cos2y}{\sin2y}=\dfrac{2\sin^2y}{2\sin y\cos y}=?$$

$$\dfrac{\cos\theta}{1-\sin\theta}=\dfrac{\sin2y}{1+\cos2y}=?$$

See also : Solve $\frac{\cos x}{1+\sin x} + \frac{1+\sin x}{\cos x} = 2$,

answered Jun 30 '17 at 06:42

lab bhattacharjee

274,582

An elegant, not to mention efficient, approach. – N. F. Taussig Jun 30 '17 at 08:48

score 1 · Answer 3 · answered Jun 30 '17 at 08:53

Set $\theta=\pi/2-x$, so the left-hand side becomes $$ \frac{1+\cos x}{\sin x}+\frac{\sin x}{1-\cos x} $$ Now recall that $$ \tan\frac{x}{2}=\frac{\sin x}{1+\cos x}= \frac{1-\cos x}{\sin x} $$ so the left-hand side is actually $$ 2\cot\frac{x}{2}=2\tan\Bigl(\frac{\pi}{2}-\frac{x}{2}\Bigr)= 2\tan\Bigl(\frac{\pi}{2}+\frac{\theta}{2}-\frac{\pi}{4}\Bigr) $$

score 0 · Answer 4 · answered Jun 30 '17 at 10:42

LHS: $$\frac{(1 + \sin\theta)(1-\sin\theta)}{\cos\theta(1-\sin\theta)} + \frac{\cos\theta}{1 - \sin\theta} = \frac{2\cos\theta}{1 - \sin\theta}.$$

RHS: $$2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)=2\frac{\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}+1}{1-\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}}=2\frac{\sin\frac{\theta}{2}+\cos\frac{\theta}{2}}{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}=$$ $$2\frac{\left(\sin\frac{\theta}{2}+\cos\frac{\theta}{2}\right)\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)}{\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)^2}=\frac{2\cos\theta}{1 - \sin\theta}.$$

Trig Identity Proof $\frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = 2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)$

4 Answers4