Simplify the trigonometric expression

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Simplify the trigonometric expression

Mathematics
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@EllenJaz17 what is the expression
\[\frac{ \sin ^{2} }{ 1-}\]
its supposed to be sin^2theta/1-cos theta

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Other answers:

the answer is 1
correct @EllenJaz17
\[\frac{ \sin ^{2}\theta }{1-cos }\]
and theta is after the cos.
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\[\mathrm{Use\:the\:following\:identity}:\quad \:1=\cos ^2\left(x\right)+\sin ^2\left(x\right)\] \[=\frac{\sin ^2\left(θ\right)}{\cos ^2\left(θ\right)-\cos ^2\left(θ\right)+\sin ^2\left(θ\right)}\] \[\frac{\sin ^2\left(θ\right)}{-\cos ^2\left(θ\right)+\cos ^2\left(θ\right)+\sin ^2\left(θ\right)}\] \[\mathrm{Add\:similar\:elements:}\:-\cos ^2\left(θ\right)+\cos ^2\left(θ\right)=0\] \[=\frac{\sin ^2\left(θ\right)}{\sin ^2\left(θ\right)+0}\] \[=\frac{\sin ^2\left(θ\right)}{\sin ^2\left(θ\right)}\] =1
this is the 1 problem
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Those are my answer options
ohhhhhh you needed just answer @EllenJaz17
yeah. Those are my options A-D in the attachments
Would the answer be C?
HI!!
hey
is it just \[\huge \frac{1-\sin(x)}{\cos(x)}\]
So it's b?
no i am still trying to figure out what the original question one
was
Simplify it
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can you repost the original one
\[\mathrm{Simplify}\:\frac{1-\sin \left(θ\right)}{\cos \left(θ\right)}:\quad \left(1-\sin \left(θ\right)\right)\sec \left(θ\right)\]
\[\huge \frac{\sin^2(\theta)}{1-\cos(\theta)}\]
we can do a couple things if that is the original question
Simplify the trigonometric expression (is the question)
perhaps the easiest it to rewrite \(\sin^2(\theta)\) as \(1-\cos^2(\theta)\) then factor
Use the identity sin^2 = 1-cos^2. Then factor it into (1+cos)(1-cos). Then cancel
\[\frac{\sin^2(\theta)}{1-\cos(\theta)}=\frac{1-\cos^2(\theta)}{1-\cos(\theta)}=\frac{(1+\cos(\theta))(1-\cos(\theta))}{1-\cos(\theta)}=1+\cos(\theta)\]
Is that the final answer?
not sure if that is an answer choice, but if it is, pick that one
It is! Thank you!
\[\color\magenta\heartsuit\]

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