Q.If three angles A, B and C are in arithmetic progression, prove that $\cot B = \dfrac{\sin A - \sin C}{\cos C - \cos A}$.

AP gives $B=\dfrac{A+C}{2}$. Using $\sin A-\sin C$ and $\cos C-\cos A$ as products, the ratio becomes $\cot\dfrac{A+C}{2}=\cot B$.

This uses the sum/difference (compound-angle-based) transformation formulae from the Trigonometry chapter of the TN HSC Class-11 Business Mathematics syllabus.

Step 1 — Use the AP condition.

$A,B,C$ in AP $\Rightarrow 2B=A+C\Rightarrow B=\dfrac{A+C}{2}.$

Step 2 — Transform the numerator.

$\sin A-\sin C=2\cos\!\left(\dfrac{A+C}{2}\right)\sin\!\left(\dfrac{A-C}{2}\right).$

Step 3 — Transform the denominator.

$\cos C-\cos A=2\sin\!\left(\dfrac{A+C}{2}\right)\sin\!\left(\dfrac{A-C}{2}\right).$

(Here $\cos C-\cos A=-(\cos A-\cos C)=2\sin\frac{A+C}{2}\sin\frac{A-C}{2}$.)

Step 4 — Divide. …