Prove that: 9A- 7 A 7A- 9A = 8A - Vidyadip

Question

Prove that:
$\frac{\sin9\text{A}-\sin\text7{A}}{\cos7\text{A}-\cos9\text{A}}=\cot8\text{A}$

✓

Answer

We have,
$\text{LHS}=\frac{\sin9\text{A}-\sin7\text{A}}{\cos7\text{A}-\cos9\text{A}}$
$=\ \frac{2\sin\Big(\frac{9\text{A}-7\text{A}}{2}\Big)\cos\Big(\frac{9\text{A}+7\text{A}}{2}\Big)}{-2\sin\Big(\frac{7\text{A}+9\text{A}}{2}\Big)\sin\Big(\frac{7\text{A}-9\text{A}}{2}\Big)}$
$=\ \frac{-\sin\text{A}\cos8\text{A}}{\sin8\text{A}\sin(-\text{A})}$
$=\ \frac{-\sin\text{A}\cos8\text{A}}{-\sin\text{A}\times\sin8\text{A}}$ $[\because\ \sin(-\theta)=-\sin\theta]$
$=\ \frac{\cos8\text{A}}{\sin8\text{A}}$
$=\ \cot8\text{A}$
$=\ \text{RHS}$
$\therefore\ \frac{\sin9\text{A}-\sin7\text{A}}{\cos7\text{A}-\cos9\text{A}}=\cot8\text{A}.$ Hence proved.

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