(1_ cos ^2)(1_tan^2)=tan^2

Answer 1

really? This is just
(sin^2)(sec^2)
need I say more?

and that's -, not _

Answer 2

Assuming you meant:
(1 - cos^2 θ)(1 -tan^2 θ) = tan^2 θ

Secondly, you don't say what to do with this, I will assume
you want to solve, since it is not an identity

Thirdly, you didn't state a domain , so I will assume 0 ≤ x ≤ 2π

1 - tan^2 θ - cos^2 θ + (cos^2 θ)(tan^2 θ) = tan^2 θ
1 - sin^2 θ/cos^2 θ - cos^2 θ + sin^2 θ = sin^2 θ/cos^2 θ
multiply by cos^2 θ
cos^2 θ - sin^2 θ - cos^4 θ + (1 - cos^2 θ)(cos^2 θ) = 1 - cos^2 θ
cos^2 θ - 1 + cos^2 θ - cos^4 θ + cos^2 θ - cos^4 θ = 1 - cos^2 θ
-2cos^4 θ + 4cos^2 θ - 2 = 0
cos^4 θ - 2cos^2 θ + 1 = 0
(cos^2 θ - 1) = 0
cos^2 θ = 1
cosθ = ± 1
θ = 0, π, 2π

pwhewww!

Answer 3

Argghhh,
always trust oobleck to cut through the fog.

Answer 4

Assuming you mean (1 - cos^2 θ)(1 -tan^2 θ) = tan^2 θ:

There are two Pythagorean Identities that can be applied here, 1+cos^2 θ = sin^2 θ, and 1 - tan^2 θ = sec^2 θ.

Thus, we have (sin^2 θ)(sec^2 θ) = tan^2 θ

Since sec^2 θ = 1/cos^2 θ, then we have sin^2 θ/cos^2 θ = tan^2 θ and they are indeed equal to each other by the Quotient Identity.

Answer 5

Correction: 1+tan^2θ = sec^2θ

Answer 6

But the actual question contained
1 - tan^2 θ which is NOT equal to sec^2 θ

so my solution is right after all,
The 1+tan^2θ was only found in the title, not in the question