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anonymous

  • one year ago

Cosx/(1+sinx)+(1+sinx)/cos(x)=2sec(x) I don't know how this is done at all and its so frustrating -__-

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  1. mathstudent55
    • one year ago
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    \(\dfrac{\cos x}{1 + \sin x} + \dfrac{1 + \sin x}{\cos x} = 2 \sec x\) \(\dfrac{\cos x}{1 + \sin x} \times \dfrac{\cos x}{\cos x} + \dfrac{1 + \sin x}{\cos x} \times \dfrac{1 + \sin x}{1 + \sin x}= 2 \sec x\) \(\dfrac{\cos^2 x}{(1 + \sin x) \cos x} + \dfrac{(1 + \sin x)^2}{(1 + \sin x) \cos x } = 2 \sec x\) \(\dfrac{\color{red}{\cos^2 x} + 1 + 2 \sin x + \color{red}{\sin^2 x}} {(1 + \sin x) \cos x } = 2 \sec x\) \(\dfrac{1+ 2 \sin x + \color{red}{1}} {(1 + \sin x) \cos x } = 2 \sec x\) \(\dfrac{2 + 2 \sin x}{1 + \sin x)\cos x} = 2 \sec x\) \(\dfrac{2\cancel{(1 + \sin x)}}{\cancel{(1 + \sin x)}\cos x} = 2 \sec x\) \(\dfrac{2}{\cos x} = 2 \sec x\) \(2 \sec x = 2 \sec x\)

  2. mathstudent55
    • one year ago
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    Start simply by finding the LCD of the two fractions on the left sides and adding the fractions together. Then simplify using the identity \(\sin ^2 x + \cos ^2 x = 1\) shown in red above.

  3. anonymous
    • one year ago
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    @mathstudent55 you are soo amazing! You seriously made it look so easy lol I appreciate you taking the time to figure it out!!

  4. mathstudent55
    • one year ago
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    You are very welcome. I was glad to be of help.

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