A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
cot x sec^4x = cot x + 2 tan x + tan^3x
this is the problem I need to simplify. I'm not certain whether I can think of something like 2 tan as tan^2. I'm also having problems finding identities that will help me. Thank you so much!
 2 years ago
cot x sec^4x = cot x + 2 tan x + tan^3x this is the problem I need to simplify. I'm not certain whether I can think of something like 2 tan as tan^2. I'm also having problems finding identities that will help me. Thank you so much!

This Question is Open

wio
 2 years ago
Best ResponseYou've already chosen the best response.2What happens when you sub \(\sec^2(x) = \tan^2(x)+1\)?

wio
 2 years ago
Best ResponseYou've already chosen the best response.2Since \(\sec^4(x) = [\sec^2(x)]^2\)

vittoria
 2 years ago
Best ResponseYou've already chosen the best response.0So i can use that formula, but instead of tan^2(x) + 1. I do [tan^2(x) + 1]^2?

wio
 2 years ago
Best ResponseYou've already chosen the best response.2Then foil it out and see if you've gotten closer.

vittoria
 2 years ago
Best ResponseYou've already chosen the best response.0cot x [tan^2(x) + 1]^2 = cot x + 2 tan x + tan^3x which turns into : cot x tan^4(x^2) + 1 = cot x + 2 tan x + tan^3x correct?

wio
 2 years ago
Best ResponseYou've already chosen the best response.2Not quite... \[ [\tan^2(x)+1]^2 = [\tan^4(x) + 2\tan^2(x) + 1] \]

wio
 2 years ago
Best ResponseYou've already chosen the best response.2Since \(\cot(x) = 1/\tan(x)\) it will basically lower the power of the \(\tan(x)\) terms.

vittoria
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry, I'm not sure where you got the 2 tan ^2 from
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.