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

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 1:17
• To help show your work better, here's a starting point into the instructions for how to format math formulas on this site: math.stackexchange.com/help/notation – David K Jun 30 '17 at 1:30
• 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 1:56
• Yeah I got up to that part but don't know how to get rid of the pi/4 – juA Jun 30 '17 at 1:57
• 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 2:00

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}

• Ok, I understand the first formula. How did you get the second one (after the comma)? – juA Jun 30 '17 at 1:31
• 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 2:59
• 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 3:21
• 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 6:47

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

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

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

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

• An elegant, not to mention efficient, approach. – N. F. Taussig Jun 30 '17 at 8:48

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 $$2\cot\frac{x}{2}=2\tan\Bigl(\frac{\pi}{2}-\frac{x}{2}\Bigr)= 2\tan\Bigl(\frac{\pi}{2}+\frac{\theta}{2}-\frac{\pi}{4}\Bigr)$$

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