Verify 1/1+sinx + 1/1-sinx = 2tan^2x
Combine the expression on the left side of the equation, so there is one common denominator.
ok so that would be 1-sinx/(1+sinx)(1-sinx) + 1+sinx/(1+sinx)(1-sinx) now what?
Yes, add them together, and expand the denominator.
I don't know how to expand, what exactly do I do?
We have this so far: \[\frac{1-sinx+1+sinx}{(1+sinx)(1-sinx)} = 2\tan^{2}x\]
yes
Simplify the numerator on the left side. FOIL the denominator.
so the numerator is just 2 right?
Yes.
ok so I have 2/1-sinx+sinx-sin^2x, is that right?
Yes, and that can be simplied even more.
so 1-sin2x-sin^2x?
for the denominator
Err now, the sinx 's cancel out. You should get 1/(1-sin^2x) Which simplies the 1/cos^2x
\[\frac{2}{1-\sin^2 x} = 2\tan^2x\] \[\frac{2}{\cos^2 x} = 2\tan^2x\]
alright so now what? and by the way thanks for all the help.
I'm trying to think. We don't want to mess with the 2tan^2x because we're proving that.
Hmm, this doesn't seem right at all...
2 sec^2 x = 2 tan^2 x Hmm, did you type it out right?
ahh i'm stupid sorry it was 2+2tan^2x
Hmm, ok, give me a minute.
was all this for nothing or do we have to start over?
Oh, wait this is easy lol. 2 sec^2 x 2*(1 + tan^x) 2 + 2 tan^2 x :)
is that the entire thing or is that ending from somewhere before?
No, that's the entire thing. Lemme write it out nicely.
\[2 \sec^2 x = 2 + 2\tan^2x\] \[2(1+\tan^2x) = 2 + 2\tan^x\] \[2 + 2\tan^2x = 2 + 2\tan^2x\]
how are you getting the 2sec^2x?
Remember the first half that I did, and I told you I was confused?
\[\frac{2}{\cos^2 x} = 2\sec^2x = 2+2\tan^x\]
mhm, so what should I erase? i need to show all the steps
tan^2 x
ahh ok that makes sense
Don't erase anything. I fully proved it, you just forgot to add that 2 to the right side lol.
Alright, thank you very much. I have like 4 other questions so should I make a new question or put it here? If you're willing to help me.
Um, you should make a new question. I might help, not sure lol. Sorry.
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