# Thread: Simplify

1. ## Simplify

Could someone simplify these plz
A) 5cosx >>* sin^2x - sinxcosx
sin^2x >> * sin^2x cos^2x

B) (tanx + 2)(tanx - 3) - (6 - tanx) + 2 tanx

C) cot - 1
secx - tanx

x = theta

2. B) (tanx + 2)(tanx - 3) - (6 - tanx) + 2 tanx

I believe you first need to open up the parentheses, so:
(tanx)^2 - 3tanx + 2tanx -6
combine all the like terms, and you should get:
(tanx)^2 + 2tanx - 12

3. Hello, mortyr!

The first two are just "algebra": multiplying, cancelling, and combining.

$\displaystyle C)\;\;\frac{\cot x}{\sec x - \tan x} - 1$

We have: .$\displaystyle \frac{\cot x - \sec x + \tan x}{\sec x - \tan x} \;=\;\frac{\frac{\cos x}{\sin x} - \frac{1}{\cos x} + \frac{\sin x}{\cos x}}{\frac{1}{\cos x} - \frac{\sin x}{\cos x}}$

Multiply top and bottom by $\displaystyle \sin x\cos x\!:\;\;\frac{\sin x\cos x}{\sin x\cos x}\cdot\frac{\frac{\cos x}{\sin x} - \frac{1}{\cos x} + \frac{\sin x}{\cos x}}{\frac{1}{\cos x} - \frac{\sin x}{\cos x}}$

and we have: .$\displaystyle \frac{\cos^2x - \sin x + \sin^2x}{\sin x - \sin^2x}\;=\;\frac{1 - \sin x}{\sin x(1 - \sin x)}$

Reduce: .$\displaystyle \frac{1}{\sin x} \;=\;\boxed{\csc x}$

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Here's another approach . . .
. . guarenteed to impress/surprise/terrify your teacher.

Multiply top and bottom of the fraction by $\displaystyle \sec x + \tan x\!:$

. . $\displaystyle \frac{\cot x}{\sec x -\tan x}\cdot\frac{\sec x + \tan x}{\sec x + \tan x} - 1 \;=\;\frac{\cot x(\sec x + \tan x)}{\sec^2x - \tan^2x} - 1$

Since $\displaystyle \sec^2x - \tan^2x\:=\:1$, we have: .$\displaystyle \cot x(\sec x + \tan x) - 1$

Then we have: .$\displaystyle \cot x\sec x + \cot x\tan x - 1$

Since $\displaystyle \cot x\tan x = 1$, we have: .$\displaystyle \cot x\sec x + 1 - 1 \;=\;\cot x\sec x$

And we have: .$\displaystyle \frac{\cos x}{\sin x}\cdot\frac{1}{\cos x}\;=\;\frac{1}{\sin x}\;=\;\csc x$