Few Homework Problems [Pre-Calculus]

[Foton]

I have a few homework problems for my review in Pre-Calculus for my upcoming final in the upcoming days. however, I am having trouble with only a few problems out of the several hundred we were given. Since he has no key, I am not sure how to go about these problems. Can anyone help me? Thanks!

1. Verify: (secx) - (sin^2x)(secx) = cosx

2. Rewrite 2?3(cos? + isin?) * ?3(cos?/4+ isin?/4) in standard complex (a+bi) form.

3. If the ratio of cotx = -12/5 and 3?/2 < x < 2?, find the following ratios:
sin x =
cos x =
tan x =



I'm just stuck on these three problems. Help appreciated, thanks! Once again, this is pre-calculus.

 

[BigGlenntheHeavy]

\(\displaystyle No. \ 1: sec(x)-sin^{2}(x)sec(x) \ = \ cos(x)\)

\(\displaystyle sec(x)[1-sin^{2}(x)] \ = \ cos(x)\)

\(\displaystyle Can \ you \ take \ it \ from \ here?\)

 

[Deleted member 4993]

1. Verify: (secx) - (sin^2x)(secx) = cosx

2. Rewrite 2?3(cos? + isin?) * ?3(cos?/4+ isin?/4) in standard complex (a+bi) form.

\(\displaystyle 2\sqrt{3}[\cos(\pi) + i \sin(\pi)] \cdot \sqrt{3}[\cos(\frac{\pi}{4}) + i \sin(\frac{\pi}{4})]\)

\(\displaystyle = 6[\cos(\pi) + i \sin(\pi)] \cdot [\cos(\frac{\pi}{4}) + i \sin(\frac{\pi}{4})]\)

\(\displaystyle = 6[e^{i\pi}] \cdot [e^{i\frac{\pi}{4}}]\)

Now finish it....

3. If the ratio of cotx = -12/5 and 3?/2 < x < 2?, find the following ratios: x is in fourth quadrant - where only cos(x) is positive
sin x = 1/?[1 + cot[sup:2vv211oz]2[/sup:2vv211oz](x)]

cos x = ?[1 - sin[sup:2vv211oz]2[/sup:2vv211oz](x)]

tan x = 1/cot(x)