I am stuck on these last three:

Question #1

$\displaystyle

\frac{tan^2+1}{sec(x)csc(x)}

$

**My Answer: **

-tan^2(x)+1= sec^2(x).

-sec/csc= tan (x) ... ok

**Question#2**

$\displaystyle 3cot^2(x)=1$

$\displaystyle

0\leq x \leq 2 \pi

$

Solve the following for x

**My Answer**:

-Divide both sides by 3

- take the sq. root

-cot(x)=sqrt3

-tan of sqrt3 = pi/3

So is pi/3 my answer?

$\displaystyle \textcolor{red}{\cot{x} = \pm \frac{1}{\sqrt{3}}}$

4 solutions. __Question#3. __
Rearrange the following equation to give an expression for dy

dx in terms of x and y.

$\displaystyle

1-3xy^2\frac{dy}{dx}-6x^2+\frac{dy}{dx}=5xy\frac{dy}{dx}-3x+6x^2\frac{dy}{dx}

$

**My Answer:**
Lost! Where do I begin?!

get every term with dy/dx on the same side , then factor out the dy/dx