Question

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

Answer 1

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

Answer 2

To simplify the expression (9sin^2a) / (3 - 3sin^2a), we can factor out 3 from the denominator:

(9sin^2a) / (3 - 3sin^2a) = 3sin^2a / (3(1 - sin^2a))

Next, we can use the Pythagorean identity: sin^2a + cos^2a = 1, to replace sin^2a with cos^2a:

sin^2a = 1 - cos^2a

Now, let's substitute this into the expression:

3sin^2a / (3(1 - sin^2a)) = 3(1 - cos^2a) / (3(1 - (1 - cos^2a)))

Simplify further:

3(1 - cos^2a) / (3(1 - (1 - cos^2a))) = (3 - 3cos^2a) / (3(1 - 1 + cos^2a))

Again, simplify:

(3 - 3cos^2a) / (3(1 - 1 + cos^2a)) = (3 - 3cos^2a) / (3cos^2a)

Finally, we can cancel out the common factor of 3:

(3 - 3cos^2a) / (3cos^2a) = (1 - cos^2a) / cos^2a

Now, using the Pythagorean identity again (sin^2a + cos^2a = 1), we know that cos^2a = 1 - sin^2a:

(1 - cos^2a) / cos^2a = (1 - (1 - sin^2a)) / (1 - sin^2a) = sin^2a / (1 - sin^2a)

Therefore, the simplified expression is: sin^2a / (1 - sin^2a)