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Math Help - identity

  1. #1
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    identity

    1. prove the following identity: \frac{1+cosx}{sinx}+\frac{sinx}{1+cosx}=\frac{2}{s  inx}

    2. prove the following identity: sec^2x-2sec x cos x+ cos^2x=tan^2x-sin^2 x
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  2. #2
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    Hello william
    Quote Originally Posted by william View Post
    1. prove the following identity: \frac{1+cosx}{sinx}+\frac{sinx}{1+cosx}=\frac{2}{s  inx}
    \frac{1+\cos x}{\sin x}+\frac{\sin x}{1+\cos x} = \frac{(1+\cos x)^2+\sin^2x}{\sin x (1+\cos x)}

    =\frac{1+2\cos x +\cos^2x+\sin^2x}{\sin x(1+\cos x)}

    =\frac{1+2\cos x +1}{\sin x(1+\cos x)}

    =\frac{2(1+\cos x)}{\sin x(1+\cos x)}

    =\frac{2}{\sin x}

    2. prove the following identity: sec^2x-2sec x cos x+ cos^2x=tan^2x-sin^2 x
    \sec^2x-2\sec x \cos x+ \cos^2x=(\sec x -\cos x)^2

    = \left(\frac{1}{\cos x}-\cos x\right)^2

    = \left(\frac{1-\cos^2x}{\cos x}\right)^2

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

    = \frac{(\sin^2x)(1-\cos^2x)}{\cos^2 x}

    = \frac{(\sin^2x-\sin^2x\cos^2x)}{\cos^2 x}

    =\frac{\sin^2x}{\cos^2x}-\sin^2x

    =\tan^2x -\sin^2x


    Grandad
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