Asked by stacey graham
(tanx+cotx)over(tanx-cotx)=(1) over sin^2x-cos^2x)
Answers
Answered by
Reiny
Writing everything in terms of sine and cosine usually works, so ...
LS = (sinx/cosx + cosx/sinx)/(sinx/cosx - cosx/sinx)
= [(sin^2x + cos^2x)/sinxcosx]/[(sin^2x - cos^2x)/(sinxcosx)]
= [(sin^2x + cos^2x)/sinxcosx][(sinxcosx0)/(sin^2x - cos^2x)]
= 1/(sin^2x - cos^2x)
= RS
LS = (sinx/cosx + cosx/sinx)/(sinx/cosx - cosx/sinx)
= [(sin^2x + cos^2x)/sinxcosx]/[(sin^2x - cos^2x)/(sinxcosx)]
= [(sin^2x + cos^2x)/sinxcosx][(sinxcosx0)/(sin^2x - cos^2x)]
= 1/(sin^2x - cos^2x)
= RS
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.