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Why is the answer for the following expression $\cot\theta\sin\theta$, not $\tan\theta\sin\theta$?

$\dfrac{1}{\cos\theta}-\cos\theta$ is equal to which of the following? \begin{alignat*}{2} & (1) \tan\theta\sin\theta\quad? \hspace{8em} && (3) \cos\theta\cot\theta \\ & (2) \cot\theta\sin\theta\quad\checkmark \hspace{8em} && (4) \sec\theta\sin\theta \end{alignat*}

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closed as off-topic by Aqua, GNUSupporter 8964民主女神 地下教會, Morgan Rodgers, eranreches, Brian Borchers Dec 22 '17 at 20:23

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  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Dec 22 '17 at 18:17
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    $\begingroup$ Because the book (or who wrote the checkmark) made a mistake, apparently. $\endgroup$ – egreg Dec 22 '17 at 18:18
  • $\begingroup$ Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. I have edited your question to reflect this principle. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 22 '17 at 18:34
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we have $$\frac{1}{\cos(x)}-\cos(x)=\frac{1-\cos^2(x)}{\cos(x)}=\frac{\sin^2(x)}{\cos(x)}=\tan(x)\sin(x)$$ since $$\frac{\sin(x)}{\cos(x)}\cdot \sin(x)=\tan(x)\sin(x)$$ better?

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  • $\begingroup$ Why....................... $\endgroup$ – Aqua Dec 22 '17 at 18:12
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The correct answer is $\tan\theta\sin\theta$. Who wrote the checkmark was wrong.

Indeed, for $\theta=\pi/3$ we have $$ \cot\theta\sin\theta=\frac{1}{\sqrt{3}}\frac{\sqrt{3}}{2}=\frac{1}{2} $$ but $$ \frac{1}{\cos\theta}-\cos\theta=2-\frac{1}{2}=\frac{3}{2} $$ By the way, $\cot\theta\sin\theta=\cos\theta$, so the equality $$ \frac{1}{\cos\theta}-\cos\theta=\cot\theta\sin\theta $$ only holds for $\cos^2\theta=1/2$, that is, for $$ \theta=\frac{\pi}{4}+k\frac{\pi}{2} \qquad \text{$k$ any integer} $$

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