Proving a relationship between related angles

Question

I am writing a physics paper relating the motion of objects using Loedel diagrams and standard trigonometry.

In one section, I am able to demonstrate a relationship between two related angles of motion; however, a universal proof of this relationship eludes me.

It will be vastly better to present a simple mathematical proof of this relationship, rather than examples that demonstrate it to be true.

Thus, I am seeking help for the following:

Prove that for any angle $\theta$ with a value $0<\theta<90$:

$$tan(\theta) = cot((90-\theta)/2)-1/cos(\theta)$$

I am grateful for any help.

Answer 1

With $$\sqrt{2}\sin(\dfrac{\pi}{4}-x)=\cos x-\sin x$$ $$\sqrt{2}\cos(\dfrac{\pi}{4}-x)=\cos x+\sin x$$ for easy calculation let $\dfrac{\theta}{2}=x$ then \begin{align} \cot\dfrac{90-\theta}{2}-\dfrac{1}{\cos\theta} &= \dfrac{\sqrt{2}\cos(\dfrac{\pi}{4}-\dfrac{\theta}{2})}{\sqrt{2}\sin(\dfrac{\pi}{4}-\dfrac{\theta}{2})}-\dfrac{1}{\cos\theta} \\ &= \dfrac{\cos x+\sin x}{\cos x-\sin x}-\dfrac{1}{\cos2x} \\ &= \dfrac{\cos x+\sin x}{\cos x-\sin x}-\dfrac{1}{\cos^2x-\sin^2x} \\ &= \dfrac{2\sin x\cos x}{\cos^2x-\sin^2x} \\ &= \dfrac{\sin2x}{\cos2x} \\ &= \tan2x \\ &= \tan\theta \end{align}

Answer 2

Because $$\cot\left(45^{\circ}-\frac{\theta}{2}\right)-\frac{1}{\cos\theta}=\frac{1+\tan\frac{\theta}{2}}{1-\tan\frac{\theta}{2}}-\frac{1}{\cos\theta}=$$ $$=\frac{\cos\frac{\theta}{2}+\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}-\frac{1}{\cos\theta}=\frac{\left(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\right)^2}{\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}}-\frac{1}{\cos\theta}$$ $$=\frac{1+2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{\cos\theta}-\frac{1}{\cos\theta}=\frac{\sin\theta}{\cos\theta}=\tan\theta$$

Answer 3

Rearranging and using that $\,\;1+\sin(\theta)=2 \sin^2(\pi/4+\theta/2)\;$ (which follows from the general sum of sines identity $\sin(a)+\sin(b)=2\sin\big((a+b)/2\big)\cos\big((a-b)/2\big)\,$ with $a=\pi/2, b=\theta\,$):

$$\require{cancel} \begin{align} \tan(\theta) + \frac{1}{\cos(\theta)} &= \frac{1+\sin(\theta)}{\cos(\theta)} \\[5px] &= \frac{2 \sin^2\left(\pi/4+\theta / 2\right)}{\sin(\pi/2 + \theta)} \\[5px] &= \frac{\cancel{2} \sin^\bcancel{2}\left(\pi/4+\theta / 2\right)}{\cancel{2} \bcancel{\sin\left(\pi/4+\theta / 2\right)}\cos\left(\pi/4+\theta / 2\right)} \\[5px] &= \tan(\pi/4+\theta/2) \\[5px] &= \cot(\pi/4-\theta/2) \end{align} $$