Asked by Rosh
[1 + sin(theta) + cos(theta)] / [1 + sin(theta) - cos(theta)] = [1+cos(theta)]/ sin(theta)
Answers
Answered by
Reiny
Suggestion:
On LS, multiply top and bottom by (1 + sinθ + cosθ)/(1 + sinθ + cosθ)
Be careful, you will have 9 terms on the top and 9 at the bottom
On the top you will have among them sin^2 θ + cos^2 θ, replace that with 1 to get
on top: 2 + 2sinθ + 2cosθ + 2sinθcosθ
group it and factor the top to 2(1+sinθ)(1+cosθ)
on the bottom: replace 1 with sin2 θ + cos^2 θ, that way the +cos^2 θ will cancel the -cos^2 θ at the end
other terms will cancel as well, leaving you with
= 2sin^2 θ + 2sinθ
= 2sin(sinθ + 1)
Well, well, will you look at that!!!
On LS, multiply top and bottom by (1 + sinθ + cosθ)/(1 + sinθ + cosθ)
Be careful, you will have 9 terms on the top and 9 at the bottom
On the top you will have among them sin^2 θ + cos^2 θ, replace that with 1 to get
on top: 2 + 2sinθ + 2cosθ + 2sinθcosθ
group it and factor the top to 2(1+sinθ)(1+cosθ)
on the bottom: replace 1 with sin2 θ + cos^2 θ, that way the +cos^2 θ will cancel the -cos^2 θ at the end
other terms will cancel as well, leaving you with
= 2sin^2 θ + 2sinθ
= 2sin(sinθ + 1)
Well, well, will you look at that!!!
Answered by
Rosh
Thanks for teaching me how to solve my trig problems. I had trouble with those two and your explanations guided me how to solve these types of problems. :)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.