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I am looking at a worked out answer to a problem I got wrong. Part of the work shows this simplification:

$=2\cdot\csc(x)\cdot\sec(x)+2x\cdot−\csc(x)\cot(x)\cdot\sec(x)+2x\cdot\csc(x)\cdot\sec(x)\tan(x)$ $=2\csc(x)\sec(x)−2x\csc^2(x)+2x\sec^2(x)$

There's no explanation of how they got from the first expression to the second, and I can't figure it out.

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Hint

Substitute $\tan(x)$ and $\cot(x)$ :

$$\tan(x)=\frac {\sec(x)}{\csc(x)} \text{, and }\cot(x)=\frac {\csc(x)}{\sec(x)}$$

Edit for Nope

$$\tan(x)=\frac {\sin(x)}{\cos(x)}=\frac 1 {\cos(x)}\sin(x)=\frac {sec(x)}{\csc(x)}$$

And then conclude for $\cot(x)$: $$\cot(x)=\frac 1 {\tan(x)}=\frac {csc(x)}{\sec(x)}$$

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    $\begingroup$ In searching for lists of trig identities, I find these ones are often unlisted. Your way is better, but it made me realize that there is also a way to do this by using more typical trig identities by putting everything in terms of sin and cos. $\endgroup$ – nope Jan 26 '18 at 4:13
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    $\begingroup$ Well Nope take the definition of $tan(x)$ it's equal to $ \frac {sin x}{\cos(x)}$ then remember $csc(x)$ is inverse of $\sin$ function and $\sec(x)$ is inverse of $\cos(x)$ simply then you get $\tan(x)=\frac {sec(x)}{csc(x)}$ $\endgroup$ – Isham Jan 26 '18 at 4:15
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Consider the definitions:

csc $\equiv \frac{\text{hypotenuse}}{\text{opposite}}$, sec$ \equiv \frac{\text{hypotenuse}}{\text{adjacent}}$, cot$ \equiv \frac{\text{adjacent}}{\text{opposite}}$

Now, it is easy to see how the second term is simplified, i.e., how $-2x\csc(x)\cot(x)\sec(x)$ is equivalent to $-2x\csc^2(x)$, if we allow for a little impropriety:

$-2x\frac{\text{hypotenuse}}{\text{opposite}}\frac{\text{adjacent}}{\text{opposite}} \frac{\text{hypotenuse}}{\text{adjacent}}= -2x\frac{\text{hypotenuse}^2}{\text{opposite}^2} \equiv-2x\csc^2(x)$

The third term is easily simplified in a similar manner.

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