# Simplify

• Jan 21st 2007, 07:36 AM
mortyr
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
• Jan 21st 2007, 08:05 AM
turtle
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
• Jan 21st 2007, 08:16 AM
Soroban
Hello, mortyr!

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

Quote:

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

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

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

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

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

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

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

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

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

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

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

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