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Factor the expression and use the fundamental identities to simplify. There is more than one correct form of the answer. $$7 \sin^2 x \csc^2 x − 7 \sin^2 x$$

I'm reviewing for a test and going over my old homework, is 7 a possible solution (I'm asking because the last time I did this on my homework I got $7\cos(x)$?

$7 \sin^2 x (csc^2 x − 1)$

$=7 \sin^2 x (1 + \cot^2x)$

$=7 \sin^2 x (1 + \frac {\cos^2x}{\sin^2x})$

$=7 \sin^2 x (\frac{\sin^2x}{\sin^2x} + \frac {\cos^2x}{\sin^2x})$

$=7\sin^2x (\frac{\sin^2x+\cos^2x}{\sin^2x})$

$=7\sin^2x (\frac{1}{\sin^2x})$

$=\frac{7\sin^2x}{\sin^2x}$

$=7$

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    $\begingroup$ $\csc ^{ 2 }{ x } -1=\cot ^{ 2 }{ x } \\ $ $\endgroup$ – haqnatural Aug 5 '15 at 21:39
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    $\begingroup$ Does the given expression equal $7$ if $x = 0$? $\endgroup$ – anomaly Aug 5 '15 at 21:40
  • $\begingroup$ use \ before \sin \cos $\endgroup$ – 3SAT Aug 5 '15 at 21:40
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    $\begingroup$ Notice that $\sin^2x\csc^2x=1$ $\endgroup$ – user84413 Aug 5 '15 at 21:41
  • $\begingroup$ @anomaly Tried to calculate $\csc^2 0$, my computer exploded. $\endgroup$ – peterwhy Aug 5 '15 at 21:43
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$$7\sin^2 x\csc^2 x-7\sin^2 x=7\sin^2 x\frac{1}{\sin^2x}-7\sin^2 x=$$ $$=7-7\sin^2 x=7(1-\sin^2 x)=7\cos^2x$$

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  • $\begingroup$ This is what should come to mind first. Why does anyone need to factor out the $7\sin^2x$ first? $\endgroup$ – G-man Jan 17 '16 at 10:19
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Notice, your mistake $\csc^2-1\ne 1+\cot^2 x$, one can easily simplify as follows $$7\sin^2 x\csc^2 x-7\sin^2 x$$ $$=7\sin^2 x(\csc^2 x-1)$$ $$=7\sin^2 x(\cot^2 x)$$ $$=7\sin^2 x\left(\frac{\cos^2 x}{\sin^2 x}\right)$$ $$=\color{red}{7\cos^2x}$$

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