Question

help???
root1-sin^2x/tan^2x+1??

is it cos^2x??

Answer 1

\[\dfrac{\sqrt{1 - \sin^2 x}}{\tan^2 x + 1} = \dfrac{\sqrt{\cos^2x}}{\sec^2 x} = \dfrac{\cos x}{\sec^2 x}\]

Answer 2

Well, actually, \(\sqrt{\cos^2 x} = |\cos x|\)... just assumed cos(x) for simplicity.

Answer 3

Now, can you complete it?

Answer 4

that right but u can divide an get tan(x/2)=1

Answer 5

Hmm?

Answer 6

then it would be 1?...
these are the answer choices!
0

1

cos^2θ

cot^2θ

Answer 7

yea 1

Answer 8

Wait, is that \(\sqrt{1 - \sin^2x}\) or \(1 - \sin^2x\)?

Answer 9

its squared both numerator & denominator!

Answer 10

Is this the question?\[\dfrac{1 - \sin^2 x}{\tan^2x + 1}\]

Answer 11

with the squareroot covering the whole thing at once!

Answer 12

Ah! Sorry, lol.

Answer 13

\[\dfrac{\cos^2 x}{\sec^2 x } = \cos^4 x\]The root of that is?

Answer 14

cos^2x?

Answer 15

That's it!

Answer 16

yay! then i was correct afterall! (: