Question

9sin^2a / 3-3 sin^2 a

Answer 1

$\frac{9sin^2 \alpha }{3-3 sin^2 \alpha}=\frac{9sin^2 \alpha }{3(1-sin^2 \alpha)}=\frac{3sin^2 \alpha }{cos^2 \alpha}=3tg^2 \alpha$