1. ## need homework help-trig

I need to simplify these equations:

(1-cosx)(1+secx)cosx

tanx+cotx/sec^2x

(tan^2x/secx+1)+1

And for these, I need to solve for the degrees (nearest tenth)

tan^2x=1

2tan^2x=3tanx-1

3sinx=cosx

secx=2cscx

Any help is greatly appreciated!!!

2. Hello, legolewinsky!

Here are the first few . . .

$\displaystyle (1-\cos x)(1+\sec x)\,\cos x$
We have: . $\displaystyle (1 - \cos x)\underbrace{\cos x\left(1 + \frac{1}{\cos x}\right)}$

. . . . . . $\displaystyle = \quad(1-\cos x)\quad(\cos x + 1) \;\;=\;\;1-\cos^2\!x \;\;=\;\;\sin^2\!x$

$\displaystyle \frac{\tan x+\cot x}{\sec^2\!x}$
We have: . $\displaystyle \frac{\tan x + \dfrac{1}{\tan x}}{\sec^2\!x} \;\;=\;\;\frac{\dfrac{\overbrace{\tan^2\!x+1}^{\te xt{This is }sec^2\!x}}{\tan x}}{\sec^2\!x} \;\;=\;\;\frac{\dfrac{\sec^2\!x}{\tan x}}{\sec^2\!x} \;\;=\;\;\frac{1}{\tan x}\;\;=\;\;\cot x$

$\displaystyle \frac{\tan^2\!x}{\sec x+1}+1$
Since $\displaystyle \tan^2\!x \:=\:\sec^2\!x-1$, we have: . $\displaystyle \frac{\sec^2\!x - 1}{\sec x + 1} + 1$

Factor and reduce: . $\displaystyle \frac{(\sec x - 1)(\sec x + 1)}{\sec x + 1} + 1 \;\;=\;\;(\sec x - 1) +1 \;\;=\;\;\sec x$

3. ## thanks

Thank you so much!