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.



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 $\cot(x)$: $$\cot(x)=\frac 1 {\tan(x)}=\frac {csc(x)}{\sec(x)}$$

  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.