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Mathematics 20 Online
OpenStudy (anonymous):

solve the trig identities in a single trig function: could you help me!!?? http://bamboodock.wacom.com/doodler/50e89ae1-6dbc-46c5-b145-b878b601ef5a

OpenStudy (anonymous):

the mistake here is that \(1-\cos(x)\neq \sin(x)\)

OpenStudy (ajprincess):

\(1-\cos^2\theta=\sin^2\theta\) \(\sin\theta=\sqrt{1-\cos^2\theta}\)

OpenStudy (anonymous):

you actually have to add the fractions first

OpenStudy (anonymous):

\[\frac{1}{1+\cos(x)}+\frac{1}{1-\cos(x)}=\frac{1+\cos(x)+1-\cos(x)}{(1+\cos(x))(1-\cos(x))}\]

OpenStudy (anonymous):

now the numerator is 2, and the denominator is \(1-\cos^2(x)=\sin^2(x)\)

Parth (parthkohli):

But I see that the question doesn't say \(\dfrac{1}{1 + \cos(x)} + \dfrac{1}{1 - \cos(x)}\)

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