proving trigonometric identities (help! due tomorrow!)?

Question

1.) cot Θ + tan Θ = csc^2 Θ + sec^2 Θ / csc Θ sec Θ

that is: cot Θ + tan Θ = csc^2 Θ + sec^2 Θ (over) csc Θ sec Θ

2.) 1-2 cos Θ / sin Θ cos Θ = tan Θ - cot Θ

that is: 1-2 cos Θ (over) sin Θ cos Θ = tan Θ - cot Θ

I don't think that i copied problem 2 wrong. Maybe my professor made it that way so that the left hand side is NOT EQUAL to the right hand side.

Oh and thanks for helping guys.

Answer 1

1) Split the right into two fractions,

(csc^2 Θ + sec^2 Θ)/(csc Θ sec Θ)
= (csc^2 Θ)/(csc Θ sec Θ) + (sec^2 Θ)/(csc Θ sec Θ)
= (csc Θ)/(sec Θ) + (sec Θ)/(csc Θ)
= (1/sin Θ)/(1/cos Θ) + (1/cos Θ)/(1/sin Θ)
= (cos Θ)/(sin Θ) + (sin Θ)/(cos Θ)
= cot Θ + tan Θ

2. I think you copied this problem wrong, here's what i think it should be, start with right side.

tan Θ - cot Θ
= (sin Θ)/(cos Θ) - (cos Θ)/(sin Θ)
= (sin^2 Θ - cos^2 Θ)/(sin Θ cos Θ) (common denominator)
Now sin^2 Θ = 1 - cos^2 Θ, plug this in,
= (1 - cos^2 Θ - cos^2 Θ)/(sin Θ cos Θ)
= (1 - 2cos^2 Θ)/(sin Θ cos Θ)
(you missed a squared when you copied the problem)

Answer 2

1. I'll use x instead of Θ because it's easier to type:
[csc² x + sec² x] / [csc x sec x] =
csc² x / [csc x sec x] + sec² x / [csc x sec x] =
csc x / sec x + sec x / cscx =
cos x / sin x + sin x / cos x =
cot x + tan x

2.
tan x - cot x =
sin x / cos x - cos x / sin x =
[sin² x - cos² x] / [sin x cos x] =
[1 - cos² x - cos² x] / [sin x cos x] =
[1 - 2cos² x] / [sin x cos x]
think you have a miscopy on this one.