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I'm having a really hard time understanding how to do these. The directions are to verify that each of the following is an identity:

$$\dfrac{\csc x}{\cot x+\tan x}=\cos x$$

I have to get the left side to equal the right.

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$$\dfrac{\csc x}{\cot x+\tan x}=\dfrac{1}{\sin x\dfrac{\cos^2 x+\sin^2 x}{\sin x\cos x}}=\dfrac{1}{\dfrac{1}{\cos x}}=\cos x,\text{ since $\cos^2+\sin^2 x=1$.}$$

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$$ \frac{ \csc x }{ \cot x + \tan x} = \frac{\frac{1}{\sin x}}{\frac{\cos x}{\sin x} + \frac{ \sin x}{ \cos x}} = \frac{\frac{1}{\sin x}}{\frac{ \cos^2 x + \sin^2 x}{\sin x \cos x}} = \frac{ \frac{1}{\sin x}}{\frac{1}{\cos x \sin x}} = \frac{ \cos x \sin x}{\sin x} = \cos x$$

I have used the identiy $\sin^2 x + \cos^2 x = 1 $

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  • $\begingroup$ This may be a silly question, but how did you get from step #2 to step #3? How did you get the sin & cos squared, & the other sin & cos multiplied instead of added? $\endgroup$ – Jordan Apr 15 '14 at 1:35
  • $\begingroup$ $$ \frac{a}{b} + \frac{c}{d} = \frac{ ad + bc}{bd} $$ $\endgroup$ – user139708 Apr 15 '14 at 3:08

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