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Math Help - trig identities

  1. #1
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    trig identities

    i understand how to prove them and all
    but i just need help proving the following 2:

    1) (sinx + tanx)/cosx + 1 = tan x


    2) tanx + tan^3x = sec^2x/cot x


    thanks
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  2. #2
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    Hello, checkmarks!

    1)\;\;\frac{\sin x + \tan x}{\cos x + 1} \;= \;\tan x

    We have: . \dfrac{\sin x + \frac{\sin x}{\cos x}}{\cos x + 1}

    Multiply by \frac{\cos x}{\cos x}: . \dfrac{\cos x}{\cos x}\cdot\dfrac{\sin x + \frac{\sin x}{\cos x}}{\cos x + 1} \;=\;\frac{\sin x\cos x + \sin x}{\cos x(\cos x + 1)}

    Factor: . \frac{\sin x(\cos x + 1)}{\cos x(\cos x + 1)} \;=\;\frac{\sin x}{\cos x} \;=\;\tan x




    2)\;\;\tan x + \tan^3\!x \;= \;\frac{\sec^2\!x}{\cot x}

    We have: . (1 + \tan^2\!x)\tan x \;=\;\sec^2\!x\cdot\tan x \;=\;\sec^2\!x\cdot\frac{1}{\cot x} \;=\;\frac{\sec^2\!x}{\cot x}<br />


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  3. #3
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    thank you so much

    i have another one..last one, i promise

    #3 (1-cos^2x)(sec^2x - 1) = tan^2x
    asgaghkcos^2xagjahajaaha
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  4. #4
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    Hello, checkmarks!

    This one is not true . . . probably a typo . . .


    3)\;\;\frac{(1-\cos^2\!x)(\sec^2\!x - 1)}{\cos^2\!x} \:= \:\tan^{{\color{red}4}}\!x

    We have: . \frac{\overbrace{(1 - \cos^2\!x)}^{\text{This is }\sin^2\!x}\,\overbrace{(\sec^2\!x - 1)}^{\text{This is }\tan^2\!x}}{\cos^2\!x} \;\;=\;\;\frac{\sin^2\!x\cdot\tan^2\!x}{\cos^2\!x} \;=\;\frac{\sin^2\!x}{\cos^2\!x}\cdot\tan^2x

    . . =\;\;\left(\frac{\sin x}{\cos x}\right)^2\!\!\cdot\tan^2\!x \;\;= \;\;\tan^2\!x\cdot\tan^2\!x \;\;=\;\;\tan^4\!x

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