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Thread: Simplify

  1. #1
    mortyr
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    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
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  2. #2
    Junior Member
    Joined
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    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
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  3. #3
    Super Member

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


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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

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