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do you know
\[ \cos^2 x + \sin^2 x = 1 \] ?

uhhhh...... lol sorry I have my ditzy moments..... I think so?

yeahhhh actually I don't

ok so far i'm following

in other words, in your math career you should have seen
\[ a^2 - b^2 = (a-b)(a+b) \]

From a post up above, you know
cos^2 theta = what ?

1 - sin^2 (theta)?

a = 1 and b = sin? I think? Maybe? Possibly?

So I'm guessing it would look like 1^2 - sin(x)^2 = (1-sin(x))(1+sin(x))?

Last step. anything divided by itself is 1
you have a (1-sin x)/(1-sin x)
which "cancel"

what is left over? that is the answer

I'm leaning toward either A or D

You don't have to lean. You should look at
\[ \frac{ (1-\sin(x))(1+\sin(x))}{(1-\sin(x))} \]

OH!

I see now! Light bulb came on!

Do you understand that means 1-sin x divided by itself (leaving an extra factor of (1+sin x)

yep! ty ty ty ty ty you're awesome!!!

It is the same problem as
\[ \frac{6}{3} = \frac{3 \cdot 2 }{3} = \frac{3}{3} \cdot 2 = 2\]

definitely. ty sooooo much again!!!!!