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Best Answer

tan^2(x)

Proof:

cos^2(x)+sin^2(x)=1 (Modified Pythagorean theorem)

sin^2(x)=1-cos^2(x) (Property of subtraction)

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

sin^2(x)/cos^2(x)=? (Property of substitution)

sin(x)/cos(x) * sin(x)/cos(x) = tan(x) * tan(x) (Definition of tanget)

= tan^2(x)

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Q: What equals 1 - cos squared x divided by cos squared x?
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Related questions

What equals 1 cos squared x divided by cos squared x?

1. Anything divided by itself always equals 1.


What is Sin squared x - Cos squared x divided by 1 - Tan squared x equals cos squared x?

22


What does cos squared x - Sin squared x equal?

2 x cosine squared x -1 which also equals cos (2x)


Is 1- cos 2 x 1 plus cos 2 x equals sin squared x cos squared x an identity?

No, (sinx)^2 + (cosx)^2=1 is though


How do you prove that the sin over one minus the cosine minus one plus the cosine over the sine equals zero?

Multiply both sides by sin(1-cos) and you lose the denominators and get (sin squared) minus 1+cos times 1-cos. Then multiply out (i.e. expand) 1+cos times 1-cos, which will of course give the difference of two squares: 1 - (cos squared). (because the cross terms cancel out.) (This is diff of 2 squares because 1 is the square of 1.) And so you get (sin squared) - (1 - (cos squared)) = (sin squared) + (cos squared) - 1. Then from basic trig we know that (sin squared) + (cos squared) = 1, so this is 0.


If Sin equals x and Cos equals y then x squared equals what function of y?

If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²


24 divided by 6 - 2 squared - 1 squared equals 9?

1


What is 1 minus cos squared?

sin squared


Verify that sin minus cos plus 1 divided by sin plus cos subtract 1 equals sin plus 1 divided by cos?

[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,


What is (1 cos x)(1- cos x)?

(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)


Can you make cosine squared X divided by2 equal to cosine squared x with some constant?

You could just pull out the half: it will be (1/2) cos squared x.


How do you solve theta if cos squared theta equals 1 and 0 is less than or equal to theta which is less than 2pi?

cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0