A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
prove: sinx+cosx/tan^2x1 = cos^2x/sinxcosx
anonymous
 5 years ago
prove: sinx+cosx/tan^2x1 = cos^2x/sinxcosx

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sin ^{4}x  \cos ^{4}x/ \tan ^{4}x 1 = \cos ^2x\]

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0I'm going to look at the first equation for now. Is the second a different proof that you need to do? There are two things to keep in mind here. First off, remember when trying to prove these that you can work both sides (it will be easier). Secondly, remember that \(\tan x = \frac{\sin x}{\cos x}\), which means that \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\). So, once we've determined that, we can first take \(\sin x  \cos x\) from the right and multiply it into the left: \[\begin{align} \frac{\sin x + \cos x}{\tan^2 x  1} &= \frac{\cos^2 x}{\sin x  \cos x}\\ \frac{(\sin x + \cos x)(\sin x  \cos x)}{\tan^2 x  1} &= \cos^2 x\\ \frac{\sin^2 x  \cos^2 x}{\tan^2 x  1} &= \cos^2 x \end{align}\] You can then take \(\tan^2 x  1\) and multiply it into the right: \[\begin{align} \frac{\sin^2 x  \cos^2 x}{\tan^2 x  1} &= \cos^2 x\\ \sin^2 x  \cos^2 x &= \cos^2 x(\tan^2 x  1) \end{align}\] Then we use the fact that \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\) and multiply through by \(\cos^2 x\) on the right: \[\begin{align} \sin^2 x  \cos^2 x &= \cos^2 x(\tan^2 x  1)\\ &=\cos^2 x \tan^2 x  \cos^2 x\\ &= \cos^2 x\frac{\sin^2 x}{\cos^2 x}  \cos^2 x\\ \sin^2 x  \cos^2 x = \sin^2 x  \cos^2 x \end{align}\] When we expand \(\tan^2 x\), we cancel out the \(\cos^2 x\) and we are left with the same expressions on both sides, which means we have proven the identity.
Ask your own question
Sign UpFind 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.