If $a,b,c$ are in arithmetic progression, prove that... If $a,b,c$ are in an Arithmetic Progression (AP), prove that
$$\frac{\sin{c}-\sin{a}}{\cos{a}-\cos{c}}=\cot{b}$$.
I tried setting $a,b,c$ as $(a),(a+d),(a+2d)$ respectively as they are in an AP.It does not work at all. Is there any other method???
 A: The trick is let $$a=A-d$$ $$b=A$$ and $$c=A+d$$.
$$\frac{\sin{c}-\sin{a}}{\cos{a}-\cos{c}}=\frac{\sin{(A+d)}-\sin{(A-d)}}{\cos{(A-d)}-\cos{(A+d)}}$$
$$\frac{\sin{A}\cos{d}-\cos{A}\sin{d}-\sin{A}\cos{d}-\cos{A}\sin{d}}{\cos{A}\cos{d}-\sin{A}\sin{d}-\cos{A}\cos{d}-\sin{A}\sin{d}}$$
Which simplifies to $$\frac{\cos{A}}{\sin{A}}$$
$$\frac{\cos{A}}{\sin{A}}=\cot{A}$$
$$\cot{A}=\cot{b}$$
Quod erat demonstrandum.
A: Since $\frac{a+c}{2}=b$, for $d\neq2\pi k$, where $k\in\mathbb Z$  and $\sin{b}\neq0$ we obtain $$\frac{\sin{c}-\sin{a}}{\cos{a}-\cos{c}}=\frac{2\sin\frac{c-a}{2}\cos\frac{a+c}{2}}{2\sin\frac{c-a}{2}\sin\frac{a+c}{2}}=\cot{b}$$
A: Your way also works (but a little longer):
$$\frac{\sin{c}-\sin{a}}{\cos{a}-\cos{c}}=\frac{\sin{(a+2d)}-\sin{a}}{\cos{a}-\cos{(a+2d)}}=\frac{\sin{a}\cos{2d}+\cos{a}\sin{2d}-\sin{a}}{\cos{a}-\cos{a}\cos{2d}+\sin{a}\sin{2d}}=\\ 
\frac{\sin{a}(\cos{2d-1})+\cos{a}\sin{2d}}{\cos{a}(1-\cos{2d})+\sin{a}\sin{2d}}=\frac{-2\sin{a}\sin^2{d}+2\sin{d}\cos{a}\cos{d}}{2\cos{a}\sin^2d+2\sin d\sin a\cos d}= \\ 
\frac{2\sin d(-\sin a\sin d+\cos a\cos d)}{2\sin d(\cos a\sin d+\sin a\cos d)}=\frac{\cos (a+d)}{\sin (a+d)}=\cot (a+d)=\cot{b}.$$
A: $$\sin c - \sin a = 2 \cos \left( \frac{a+c}{2} \right) \sin \left (\frac {c-a}2\right) $$
$$\cos c - \cos a = 2 \sin \left( \frac{a+c}{2} \right) \sin \left(\frac {a-c}2\right) $$
Can you continue?
