Question

plz solve it it's important trigonometry

Answer 1

$Given:\frac{sin9A}{sin3A} - \frac{cos9A}{cos3A}$

$=>\frac{sin3A}{sinA}-\frac{cos3A}{cosA}$

$=>\frac{sin3A*cosA-cos3A*sinA}{sinA*cosA}$

$=> \frac{sin(3A - A)}{sinAcosA}$

$=> \frac{sin 2A}{sinAcosA}$

$=> \frac{2sinAcosA}{sinAcosA}$

$=> \boxed{2}$

Hope it helps!