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

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*}

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Aqua, GNUSupporter 8964民主女神 地下教會, Morgan Rodgers, eranreches, Brian Borchers
If this question can be reworded to fit the rules in the help center, please edit the question.

• – Shaun Dec 22 '17 at 18:17
• Because the book (or who wrote the checkmark) made a mistake, apparently. – egreg Dec 22 '17 at 18:18
• 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. – GNUSupporter 8964民主女神 地下教會 Dec 22 '17 at 18:34

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?
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 $$\theta=\frac{\pi}{4}+k\frac{\pi}{2} \qquad \text{k any integer}$$