Question

How the heck would I simplify this???
(1/cot^2(theta))+sec(theta)cos(theta)

Answer 1

Using the identities cot = 1/tan(x) and sec(x) = 1/cos(x).

Answer 2

Well you could start by recognizing that sec and cos are reciprocals and so their product is 1

Answer 3

\[\frac{ 1 }{ \cot^2 \theta } \sec(\theta)\cos(\theta) \]

Answer 4

I knew those did that, but would I have to multiply th fraction by.. tangent or something?

Answer 5

These 3 identities might help.\[\frac{1}{Cos^2θ} =Sec^2θ,~~~~~~~~~~~~~~\frac{1}{Sec^2θ} =Cos^2θ,~~~~~~~~~~Cosθ~Secθ=1\]

Answer 6

Oh, cot, my bad

Answer 7

The tangent is the reciprocal of the cotangent so 1/cot^2=tan^2

Answer 8

\[\huge\color{blue}{ \frac{1}{Cotθ} ~Secθ~Cosθ } \]
\[\huge\color{blue}{ Tanθ~Secθ~Cosθ } \]
\[USE~~~THIS~~~IDENTITY:~~~~~~~~~~~~~~~~~\huge\color{red}{ Secθ~Cosθ =1~~} \]

Answer 9

I can't see that, but one of my choices is tan^2(theta). So that's right, right? None of them have +1.

Answer 10

Yes, you are right.

Answer 11

Also, in my work, I made an error, it's cot^2θ and tan^2θ

Answer 12

So now you have reduced it to tan^2+1. You have an identity for that.

Answer 13

Yay. You're fine. I saw it. :) I have a few more. Do you want me to post it seperately?