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Math Help - Proving Trig. Identity

  1. #1
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    Proving Trig. Identity

    Hey Guyzzz,
    I have two Identities given and I have to prove either one.

    1)  \frac{1 - tan^2 x}{1 + tan^2 x} = cos 2x

    OR

    2)  sin x +sin x cot^2 x = sec x


    Please, try you best.
    Thank you.
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  2. #2
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    Quote Originally Posted by MordernWar2 View Post

    2)  sin x +sin x cot^2 x = sec x
    I get.


     \sin (x) +\sin (x) \cot^2 (x)

     \sin (x) +\sin (x) \frac{\cos^2 (x)}{\sin^2(x)}

     \sin (x) + \frac{\cos^2 (x)}{\sin(x)}

     \frac{\sin^2 (x)}{\sin(x)} + \frac{\cos^2 (x)}{\sin(x)}

     \frac{\sin^2 (x) + \cos^2 (x)}{\sin(x)}

     \frac{1}{\sin(x)}

     <br />
\csc(x)<br />
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  3. #3
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    Hello, MordernWar2!

    1)\;\;\frac{1 - \tan^2\!x}{1 + \tan^2\!x} \:=\: \cos 2x

    We have: . \frac{1-\dfrac{\sin^2\!x}{\cos^2\!x}}{1 + \dfrac{\sin^2\!x}{\cos^2\!x}}

    Multiply by \frac{\cos^2\!x}{\cos^2\!x}; \quad \frac{\cos^2\!x\left(1 - \dfrac{\sin^2\!x}{\cos^2\!x}\right)}{\cos^2\!x\lef  t(1 + \dfrac{\sin^2\!x}{\cos^2\!x}\right)}

    . . . . . . . . =\;\;\frac{\overbrace{\cos^2\!x - \sin^2\!x}^{\text{This is }\cos2x}}{\underbrace{\cos^2\!x + \sin^2\!x}_{\text{This is 1}}} \;\;=\;\; \cos2x




    2)\;\;\sin x +\sin x\cot^2\!x \:=\: \sec x
    This is not an identity . . . I suspect a typo.

    It is probably: . \sin x + \sin x\cot^2\!x \:=\: {\color{blue}\csc x}


    \text{Factor the left side: }\:\sin x\underbrace{(1 + \cot^2\!x)}_{\text{This is }\csc^2x} \;\;=\;\;\sin x\csc^2\!x

    . . . . . . . . . . . . =\;\;\underbrace{(\sin x\csc x)}_{\text{This is 1}}\csc x \;\;=\;\;csc x

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