Showing that $\dfrac{\sin \alpha-\sin \beta}{\cos \beta-\cos \alpha}=\cot \theta$ for some $\theta$ between $\alpha$ and $\beta$

Question

I am stuck on the following problem that says:

Show that $$\frac{\sin \alpha-\sin \beta}{\cos \beta-\cos \alpha}=\cot \theta$$ where $0 \lt \alpha \lt \theta \lt \beta \lt \frac{\pi}{2}$.

Can someone explain in details? Thanks in advance for your time.

Answer 1

Apply Cauchy's mean value theorem:

If $f$ and $g$ are continuous on the closed interval $[\alpha,\beta]$, differentiable on its interior $(\alpha,\beta)$, there exists $\theta \in (\alpha,\beta)$ such that $$\frac{f(\alpha)-f(\beta)}{g(\alpha)-g(\beta)}=\frac{f'(\theta)}{g'(\theta)}.$$

Answer 2

Notice that:

$$\sin(\alpha) - \sin(\beta) = 2\sin\left(\frac{\alpha-\beta}{2}\right)\cos\left(\frac{\alpha+\beta}{2}\right)$$

$$\cos(\beta) - \cos(\alpha) = 2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)$$

Their ratio is:

$$\cot(\theta) = \displaystyle\frac{\cos\left(\displaystyle\frac{\alpha+\beta}{2}\right)}{\sin\left(\displaystyle\frac{\alpha+\beta}{2}\right)},$$

where $\theta = \displaystyle\frac{\alpha+\beta}{2}.$