Need help with a couple college-level trig problems

[DL93]

1. Prove the identity: (tanx-tan3x)/(1+tanx*tan3x)=-(2tanx)/(1-tan²x)
2. Prove the identity: (cosx-sinx)/(cos³x-sin³x)=(2)/(2+sin(2x))
3. Prove the identity: cos²Bcot²B=cot²B-cos²B

I don't understand these at all, and I would really appreciate it if anyone could help me ASAP

 

[DL93]

And please explain the steps as well

 

[srmichael]

1. Prove the identity: (tanx-tan3x)/(1+tanx*tan3x)=-(2tanx)/(1-tan²x)
2. Prove the identity: (cosx-sinx)/(cos³x-sin³x)=(2)/(2+sin(2x))
3. Prove the identity: cos²Bcot²B=cot²B-cos²B

I don't understand these at all, and I would really appreciate it if anyone could help me ASAP

You obviously didn't read the FORUM GUIDELINES. We don't work that way where we simply do your work. We need to see some effort on your part so we know where you are stuck. Lucky for you it's a Friday and I'm feeling generous.

Here's #2.

\(\displaystyle \frac{cosx-sinx}{cos^3x-sin^3x}=\frac{2}{2+sin2x}\)

\(\displaystyle \frac{cosx-sinx}{(cosx-sinx)(cos^2x+sinxcosx+sin^2x)}=\frac{2}{2+sin2x}\)

\(\displaystyle \frac{1}{cos^2x+sinxcosx+sin^2x}=\frac{2}{2+sin2x}\)

\(\displaystyle \frac{1}{1+sinxcosx}=\frac{2}{2+sin2x}\)

\(\displaystyle \frac{1}{1+\frac{1}{2}(2sinxcosx)}=\frac{2}{2+sin2x}\)

\(\displaystyle \frac{2}{2+2sinxcosx}=\frac{2}{2+sin2x}\)

\(\displaystyle \frac{2}{2+sin2x}=\frac{2}{2+sin2x}\)

Now try #1 and #3 the best you can.

P.S. Well, not that generous...too lazy to explain every step.

 

[DL93]

I'm really sorry about not reading the guidlines, but I really appreciate your feedback. I did figure out #3, but I'm still confused about #1. I'll go through the steps that I've already completed: I started out by combining the tanx -tan3x= -2tanx. Next in the denominator I took 1+tanx*tan3x to get 1+tan²3x. From this point Im utterly stuck on how to match the right side since I have the numerator correct by my denominator is still off. Thank you again for the generosity, and thank you for your time.

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[srmichael]

I'm really sorry about not reading the guidlines, but I really appreciate your feedback. I did figure out #3, but I'm still confused about #1. I'll go through the steps that I've already completed: I started out by combining the tanx -tan3x= -2tanx. Next in the denominator I took 1+tanx*tan3x to get 1+tan²3x. From this point Im utterly stuck on how to match the right side since I have the numerator correct by my denominator is still off. Thank you again for the generosity, and thank you for your time.

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No way, José. You can't combine tanx and tan3x in that way. Each term has a different angle. One is "x" and the other is "3x". Here's a hint that may help:

tan3x = tan(x+2x). Now maybe the sum of two angles formula may help you proceed.