Prove the identity: (identity in comments) (:

Question

haleyelizabeth2017 · Answer

$$\frac{2~cos(2x)}{sin(2x)}=cot(x)-tan(x)$$

haleyelizabeth2017 · Answer

OH

abdullahm · Answer

We can try to write cot(x) and tan(x) in terms of sin(x) and cos(x) and see if we can get to the left hand side c:

abdullahm · Answer

$\Large cot(x) = \frac{cos(x)}{sin(x)} $

$\Large tan(x) = \frac{sin(x)}{cos(x)} $

$\Large cot(x)-tan(x)= \frac{cos(x)}{sin(x)} -\frac{sin(x)}{cos(x)} $

And subtract those two fractions... And forget it.

Check this out:

$$sin(2x) = 2sin(x)cos(x)\

cos(2x) = cos^2(x) – sin^2(x) $$

Plug those two identities into your identity in the left hand side and let's simplify :)

haleyelizabeth2017 · Answer

$$\frac{2cos(2x)}{sin(2x)}=cot(x)-tan(x)$$$$=\frac{2(cos^2x-sin^2x)}{2sin(x)~cos(x)}$$$$=\frac{cos^2x}{sin(x)~cos(x)}-\frac{sin^2x}{sin(x)~cos(x)}$$$$=\frac{cos(x)}{sin(x)}-\frac{sin(x)}{cos(x)}$$$$=cot(x)-tan(x)$$

abdullahm · Answer

First I was hoping to simplify the right hand side and then simplify the left hand side to prove the identity, but then it struck me how we can just simplify the left hand side to get our answers.

abdullahm · Answer

Yup, you got it! Great job!

haleyelizabeth2017 · Answer

LOL Thanks XD

abdullahm · Answer

It was my pleasure! :)