(tan^2θ)(cos^2θ)=1-cos^2θ?

Question

I need the answer to this problem step by step. The left side of the equation really doesn't have parenthesis. I just did that so people would know it needs to be multiplied. thanks.

Answer 1

left side

(sin^2 / cos^2)(cos^2) = sin^2

Right side

1-cos^2 = sin^2

Answer 2

Okay, so to do these types of problems just choose one side of the equation and "work" it. I chose the left side:

(tan^2θ)(cos^2θ) = ((tanθ)^2)(cos^2θ) because tangent-squared is just the square of the function.

((tanθ)^2)(cos^2θ) = (((sinθ)^2)/((cosθ)^2))(cos^2θ) because tangent(x) is sine(x) over cosine(x). We know, however that when we have (a/b)(b), the b "cancels" out, so by the same token:
(((sinθ)^2)/((cosθ)^2))(cos^2θ) = (sinθ)^2

By Pythagorean identity, we know that ((sinθ)^2) + ((cosθ)^2) = 1, so using some algebra, we also know that (sinθ)^2 = 1 - ((cosθ)^2).

Thus, we just got the right-hand part of the equation, because (sinθ)^2 = 1 - ((cosθ)^2). Q.E.D.

Btw...when you do the actual problem I don't think a big explanation is needed for each step, just make the equation flow...also i know it's hard to do it here on y answers, but be careful with the notation. I know what you meant though...anyway good luck!

Answer 3

Since tan^2θ = sin^2θ / cos^2θ
So L.H.S = sin^2θ / cos^2θ * cos^2θ = sin^2θ = 1- cos^2θ
= R. H. S.

Answer 4

(tan^2θ)(cos^2θ)=1-cos^2θ
(sin^2θ/cos^2θ)(cos^2θ) = 1-cos^2θ
sin^2θ = 1 - cos^2θ
sin^2θ + cos^2θ = 1
1 = 1

Answer 5

whoa! you speaka english??



haha jk. I did this in class today but i dont remember.