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Please Help! I have been struggling with this problem for far too long. I have tried rewriting $\tan(x)$ as $\frac{\sin(x)}{\cos(x)}$ in the denominator then simplifying the denominator, but I get stuck there.

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    $\begingroup$ What do you get when you simplify the denominator? What is $$\frac{\sin x }{\cos x} \frac{1}{\sin x}$$ $\endgroup$ – user296602 Mar 7 '16 at 20:08
  • $\begingroup$ That's exactly where I get stuck. $\endgroup$ – Dora Mar 7 '16 at 20:09
  • $\begingroup$ @Dora what is $\frac{u}{u}$? Now let $u=\sin(x)$. $\endgroup$ – Bobson Dugnutt Mar 7 '16 at 20:11
  • $\begingroup$ Do you see that the sine terms cancel? $\endgroup$ – user296602 Mar 7 '16 at 20:11
  • $\begingroup$ my hints: do u know $\tan x=\sin x/\cos x$ & $\cosec x=1/\sin x$ $\endgroup$ – Bhaskara-III Mar 7 '16 at 21:35
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$$\frac{1}{\tan x\csc x}=\frac{1}{\frac{\sin x}{\cos x}\frac1{\sin x}}=\frac{1}{\frac{1}{\cos x}}=\cos x$$

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  • $\begingroup$ How did you get from (sinx/cosx)(1/sinx) to 1/cosx? $\endgroup$ – Dora Mar 7 '16 at 20:12
  • $\begingroup$ @Dora When you multiply by something and then divide by the same thing, it is as if you didn't do it at all; therefore we say that the two things cancel. GoodDeeds saw that $\sin(x)$ was both multiplied and divided in the expression, so they cancelled. $\endgroup$ – Bobson Dugnutt Mar 7 '16 at 20:14
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$$\frac{1}{\tan(x)\csc(x)}=\frac{1}{\tan(x)\cdot\frac{1}{\sin(x)}}=\frac{1}{\frac{\tan(x)}{\sin(x)}}=\frac{\sin(x)}{\tan(x)}=\cos(x)$$

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