Simplifying Trig Equation Problems! Please Help!!?

1.
[sec(t) - cos(t)] / [3 tan(t) sin(t)]
= {[1/cos(t)] - cos(t)} / {3 [sin(t)/cos(t)] sin(t)}
= {[1/cos(t)] - [cos²(t)/cos(t)} / [3 sin²(t)/cos(t)]
= {[1 - cos²(t)]/cos(t)} / [3 sin²(t)/cos(t)]
= [sin²(t)/cos(t)] / [3 sin²(t)/cos(t)]
= [sin²(t)/cos(t)] × {cos(t) / [3 sin²(t)]}
= 1/3


2.
[sin³(t) + cos³(t)] / [1 - sin(t) cos(t)]
= [sin(t) + cos(t)] [sin²(t) - sin(t) cos(t) + cos²(t)] / [1 - sin(t) cos(t)] ….. for u³ + v³ = (u + v)(u² - uv + v²)
= [sin(t) + cos(t)] {[sin²(t) + cos²(t)] - sin(t) cos(t)} / [1 - sin(t) cos(t)]
= [sin(t) + cos(t)] [1 - sin(t) cos(t)] / [1 - sin(t) cos(t)]
= sin(t) + cos(t)


3.
[cos³(t) + sin³(t)] / [(cos(t) + sin(t)]²
= [cos(t) + sin(t)] [cos²(t) - cos(t) sin(t) + sin²(t)] / [(cos(t) + sin(t)]² ….. for u³ + v³ = (u + v)(u² - uv + v²)
= [cos²(t) - cos(t) sin(t) + sin²(t)] / [(cos(t) + sin(t)]
= {[cos²(t) + sin²(t)] - cos(t) sin(t)} / [(cos(t) + sin(t)]
= [1 - cos(t) sin(t)] / [(cos(t) + sin(t)]


4.
{sin(t) / [1 - cos(t)]} / {[1 - cos(t)] / sin(t)}
= {sin(t) / [1 - cos(t)]} × {sin(t) / [1 - cos(t)]
= sin²(t) / [1 - cos(t)]²
= [1 - cos²(t)] / [1 - cos(t)]²
= [1 + cos(t)] [1 - cos(t)] / [1 - cos(t)]²
= [1 + cos(t)] / [1 - cos(t)]

Use a calculator. When t = 30°
{sin(30°) / [1 - cos(30°)]} / {[1 - cos(30°)] / sin(30°)} = 13.92820323
[1 + cos(30°)] / [1 - cos(30°)] = 13.92820323
2sec(30°) = 2/sin(30°) = 4

Hence, {sin(t) / [1 - cos(t)]} / {[1 - cos(t)] / sin(t)} = [1 + cos(t)] / [1 - cos(t)] ≠ 2 csc(t)

1. (should be)
(sec t - cos t) / (3 tan t sin t) ,
= (1/cos t) (1 - cos^2 t) / (3 sin^2 t / cos t)
= (sin^2 t / cos t) / (3 sin^2 t / cos t)
= 1/3

2. Use the factor form a^3 + b^3 = (a + b) (a^2 - ab + b^2) :
sin^3 t + cos^3 t
= (sin t + cos t) (sin^2 t - sin t cos t + cos^2 t)
= (sin t + cos t) (sin^2 t + cos^2 t - sin t cos t)
= (sin t + cos t) (1 - sin t cos t)
So (sin^3 t + cos^3 t) / (1 - sin t cos t) -- again note the required parentheses
= sin t + cos t

3. follows from 2.

4.
(sin t / (1 - cos t) ) / ( (1 - cos t) / sin t)
= (sin t / (1 - cos t) )^2
= (2 sin (t/2 cos (t/2)/ 2 sin^2 (t/2) )^2
= (cos (t/2) / sin (t/2) )^2
= cot^2 (t/2) ??
Was there a copying error in the statement of the problem?

#1 : you have [ sin² x / cos x] / [ 3 sin² x / cos x ] = 1 / 3

#4 . you have [ sin² t +1 - 2 cos t + cos² t ] / [ ( sin t ) ( 1 - cost ) ] ≡ [ 2 - 2 cos t] / [ sin t ( 1 - cos t ] = 2 / sin t = 2 csc t

in #2 & 3 use A³ + B³ = [ A + B ] [ A² - AB + B²]