Trigonometric Identities Help! secθ - sinθ * tanθ =cosθ?

Question

secθ - sinθ * tanθ =cosθ
I have a five problem worksheet on these, and I can do them when someone is doing it with me, but when I have to do it myself I can't.

Could someone please show me how to do this problem? My teacher said not to touch the right side of the problem.

Answer 1

secθ - sinθ * tanθ =
1/cosθ - sinθ * sinθ/cosθ

= [1 - (sinθ)^2] / cosθ
= (cosθ)^2/cosθ
= cosθ

(remember: (cosθ)^2 + (sinθ)^2 = 1)

Answer 2

A first step that often helps clarify things is to replace functions like tan and sec by their definition in terms of sin and cos. Trying that on the left-hand side in this case we get

1/cosθ - sinθ(sinθ/cosθ)

With some experience you'll see that this one step has been very helpful. Both fractions have the same denominator and the squared sin function looks very promising.

Combine the two fractions into one:

(1 - sin²θ)/cosθ

Now it is fairly easy to recognize the numerator as an expression that can be replaced by applying this form of the Pythagorean identity

1 - sin²θ = cos²θ

yielding

cos²θ/cosθ

removing the common factor gives just cosθ. Which is what was to be shown. And, we only had to transform the left side.

Answer 3

take LHS:
secθ-sinθtanθ
=1/cosθ - sinθ(sinθ/cosθ)
LCM
= (1-sin^2θ)/cosθ
=cos^2θ/cosθ
=cosθ
LHS=RHS
(cos^2θ+sin^2θ=1)

Answer 4

LHS
= 1 / cos Ø - sin Ø.(sin Ø / cos Ø)
= (1 - sin² Ø) / cos Ø
= cos² Ø / cos Ø
= cos Ø
LHS = RHS
Sorry about Ø but cannot locate theta!
As a matter of interest, how do you find theta?