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 our expert's

answer on brainly

SEE EXPERT ANSWER

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

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

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

Okay, so let me try this:
did you mean:
\[Sec^2(x)/(Sec^2x - 1) = Csc^2(x)\]

yea. sorry i don't know how get it like that on my computer

okay so now workin on it

thank you so much

\[[1/Cos^2x] / [ (1/Cos^2x)-1 ] = 1/Sin^2x\]

Now let's simplify this part:
\[(1/Cos^2x) - 1\] = \[(1-Cos^2x)/Cos^2x\]

wait. how did you do that?

find the common denominator

1 = Cos^2 / Cos^2... get it?

oh . yea, sorry. i got it

\[[1/Cos^2x] * [Cos^2x/(1-Cos^2x)] = 1/Sin^2x\]

Cancel out cos^2s

\[1/(1-Cos^2x) = 1/Sin^2x\]

Cross multiply:

wait. can't you just say 1 / (sin^2x) =csc^2x

and then be done? or is that skipping a step?

yeah, but i prefer keeping everything in sin and cosine

bcuz its easy to simplify

\[Sin^2x = 1 - Cos^2x\]

See that's proved. Sin^2x + Cos^2x = 1!!!!

get it?

yes. thank you so much!

deff will. that was so helpful. thanks again!