Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

cot(theta)-tan(theta) over sin(theta) + cos(theta) equals csc(theta) - sec(theta) how do I proof this?

TriC-MathMOOC
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Let's use x instead of theta
okay
\[\frac{\cot x-\tan x}{\sin x+\cos x}=\csc x-\sec x\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Do I have the problem stated correctly?
yes :)
would I have to replace cot(x) with cos(x)/sin(x)
and tan(x) with sin(x)/cos(x)
\[(\frac{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}}{\sin x+\cos x})\times \frac{\sin x \cos x}{\sin x \cos x}=\frac{\cos ^2x-\sin ^2x}{\sin x \cos x(\sin x+\cos x)}\]
Can you get it from there?
sadly no? can you explain how I can simplify it more please?
Factor the numerator and then the cosx + sinx will cancel.
I'm still lost :(
@alejandrop95 This is not your question
how would I factor it though? I'm sorry I am new to trig.
Can you factor this: \[a^2-b^2\]
I don't think you can because it doesn't have the same coefficient
\[a^2−b^2=(a-b)(a+b)\]
\[\cos ^2x-\sin ^2x=(\cos x-\sin x)(\cos x+\sin x)\]
oh okay that makes sense. so would it be cos-sin(cos+sin)
yes
then that would cancel with the denominator correct?
Yes. sinx + cosx will cancel
which will leave 1/sin(x)+cos(x)
no
using the reciprocal identities this will equal csc(x)-sec(x) ?
|dw:1382579245755:dw|
cos(x)-sin(x)/sin(x)+cos(x0
this is where I would use the reciprocal identities?
|dw:1382579370642:dw|
the cos(x) cancel and the sin(x) cancels?
|dw:1382579509266:dw|
yay!!!! the cancel and its one,which now you are left with the csc(x) and sec(x) reciprocal identities!! Thank you!! the key idea here that I need to work on was the factoring
thank you !:)
thanks ! :)
yw

Not the answer you are looking for?

Search for more explanations.

Ask your own question