Hard Trig Identity Question!?

Question

(1-(1/cos^2x))/(1-cos^2x)

I know that the answer is 1/(cos^2x) but getting there is another question! Any help will be appreciated. This is practise for an exam i have tommorow by the way :).

Answer 1

This is
(1 - sec^2 x) / (1 - cos^2 x)
employing two trig identities,
1) 1 + tan^2 (x) = sec^2 (x)
2) sin^2 (x) + cos^2 (x) = 1

We get -tan^2 (x) / sin^2 (x)
= - 1 / cos^2 (x)

Answer 2

I think it should be -1/(cos^2x) you dont have to use any trig identities. Factor -1/(cos^2x) out to get


-1/(cos^2x)*(1 - cos^2x)/(1-cos^2x)

all thats left is -1/(cos(x))^2

Answer 3

factor out cos^2x from the bottom to get (cos^2x-1)/cos^2x inside the left brackets then cos^2x - 1 = -sin^2x. that gets -tan^2x + cos^4x+cos^2x. take cos^4x on top of the fraction to get (cos^2x-sin^2x)/cos^2x for the fraction. then add cos^2x to get 1/cos^2x

Answer 4

(1-(1/cos^2x))/(1-cos^2x)
= ( cos^2(x) - 1) / cos^2(x) (1 - cos^2(x) )
= -1 / cos^2(x).

Answer 5

N = 1 - 1 / cos² x
N = (cos²x - 1) / cos²x
D = (1 - cos²x)
N / D = (cos²x - 1) / [ cos²x.(1 - cos²x) ]
N / D = - (1 - cos²x) / [cos²x.(1 - cos²x ]
N / D = - 1 / cos²x = - sec²x

Answer 6

(1-(1/cos^2x))/(1-cos^2x)

= [ cos^2(x) - 1 ]/(1-cos^29x))

= -1/cos^2(x)

= -sec^2(x)

Answer 7

I don't know, I failed trig.