Simplify Trig Problem? Help!?

Question

1-sin^2x
(over)
sin^2x

Thanks a lot!

Also, does anyone know the Arccos of sqrt{3}/2

Answer 1

(1 - sin²x) / sin²(x)
= cos²(x) / sin²(x)
= tan²(x)

arccos sqrt(3)/2
= π/6

Answer 2

Hi. I would guess 1. The sin^2x cancels.

Second question 2 pi / 3 http://www.analyzemath.com/Inverse-Trigonometric-Functions/questions.html

Edit. I bow to the other answers on question 1.

Answer 3

1-sin^2 x is the same thing as cos^2 x
so you sub that
and now you have

cos^2x over sin^2x, which is the same thing as tan^2x
=)

and about the arcblah thing, we are just now covering that chapter and my teacher is a crazy person so i cant help you on that one, sorry!

Answer 4

1-sin^2x = cos^2x

therefore --> 1-sin^2x/sin^2x = cos^2x/sin^2x = tan^2x