sequences and series - 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???

Answer

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.