Simplify the trig expression.

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\(\Large\frac{sin^2\theta}{1-\cos\theta}\)
\(\color{blueviolet}{\Huge\sf{sin^2\theta+cos^2 \theta\equiv 1}}\)
k?

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Other answers:

yes. but that's not an option.
You are not allowed to use that ?
I'm confused. how do I use that?
http://prntscr.com/7bv1ju my choices
instead of \(\sin^2\theta\) what would you substitute?
1-cos^2? @amilapsn
\color{blueviolet}{\Huge\sf{Yes.}}
you forgot the `\(` and `\)`
then what do I do?
I did it intentionally... :D
okay XD
then use...
what's a and b?
\[\color{blueviolet}{\Huge\sf{a^2-b^2\equiv(a-b)(a+b)}}\]
just have a guess...
(I just asked you to factorize 1-cos^\theta)
right now I have 1-cos^2/1-cos (theta)
@amilapsn Can you still help? I'm confused
can you factorize \(1-\cos^2\theta\)?
how?
you know this?
\color{blueviolet}{\Huge\sf{a^2-b^2=(a-b)(a+b)}}
not really
if you simplify the r.h.s. you'll get l.h.s. and you can prove it yourself....
rhs? lhs? what's that?
rhs = right hand side lhs = left hand side
I still don't know how to do it
\[\color{blueviolet}{\Huge\sf{1-\cos^2\theta=(1-\cos\theta)(1+\cos\theta)}}\]
ok?
yes?
Now you have the answer
oh, so is the answer 1+cos theta?
yepp
\(\Huge\color{lime}{Thank~you!}\)
\[\color{blue}{\Huge\sf{You're ~Welcome!}}\]

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