Question

I have done a bazilion problems with only a few I can't seem to get. I am lost with this one. 2/(1+cos2t)=sec^2t

Answer 1

how many is a bazillion?

Answer 2

A few more than a zillion!

Answer 3

wow... that's a lot... :)

Answer 4

I know, Can you solve this problem?

Answer 5

is the problem to prove this as an identity: \(\large \frac{2}{1+cos(2t)}=sec^2t \) ???

Answer 6

Yes

Answer 7

ok... let's just work with the left side...

Answer 8

Thanks

Answer 9

Is this one stumping you too?

Answer 10

\( \large \frac{2}{1+cos(2t)} \)

=\( \large \frac{2}{1+cos(2t)} \cdot \frac{1-cos2t}{1-cos2t}\)

=\( \large \frac{2(1-cos2t)}{1-cos^2(2t)} \)

=\( \large \frac{2(1-cos2t)}{sin^2(2t)} \)

=\( \large \frac{2(1-cos2t)}{4sin^2tcos^2t} \)

=\( \large \frac{1-cos2t}{2sin^2tcos^2t} \)

=\( \large \frac{1-(cos^2t-sin^2t)}{2sin^2tcos^2t} \)

=\( \large \frac{1-cos^2t+sin^2t}{2sin^2tcos^2t} \)

=\( \large \frac{sin^2t+sin^2t}{2sin^2tcos^2t} \)

=\( \large \frac{2sin^2t}{2sin^2tcos^2t} \)

i think u may be able to do the rest from here?

Answer 11

Wow. Now I see why it was taking awhile! You're the best!

Answer 12

thanks...

Answer 13

This was a difficult problem, but I got it. Thanks again