anonymous
  • anonymous
Verify, then find which is NOT equivalent : (Picture below)
Trigonometry
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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anonymous
  • anonymous
\[\frac{ cosx }{ sinx }\]
anonymous
  • anonymous
Can you show me how to figure that out?

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anonymous
  • anonymous
sure
anonymous
  • anonymous
sorry
anonymous
  • anonymous
\[\sin(\pi/2-x)\csc\]
anonymous
  • anonymous
is this the question
anonymous
  • anonymous
because in the picture the question does not seem right
anonymous
  • anonymous
if this is the question
anonymous
  • anonymous
\[\sin(\frac{ \pi }{2 }-x)=\sin(\frac{ \pi }{ 2 })cosx-\cos(\frac{ \pi }{ 2 })sinx=1.cosx-0.sinx=cosx\]
anonymous
  • anonymous
having this in mind, let us insert it in the original expression
anonymous
  • anonymous
\[\sin(\frac{ \pi }{2 }-x)cscx=cosxcscx=cosx.\frac{ 1 }{ sinx }\]
anonymous
  • anonymous
cosx/sinx=cotx=1/tanx
anonymous
  • anonymous
done
anonymous
  • anonymous
Thank you! I have another one that maybe you could help me with?
anonymous
  • anonymous
for sure
anonymous
  • anonymous
okay, hold on let me get it..
anonymous
  • anonymous
If you could do it all in one message that would be great so I can follow along easier.
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anonymous
  • anonymous
\[\frac{ cscA-sinA }{ \cot ^{2}A }=\frac{\frac{ 1 }{sinA }-sinA }{ \frac{ \cos ^{2}A }{ \sin ^{2}A } }=\frac{ \frac{ 1-\sin ^{2}A }{sinA } }{ \frac{ \cos ^{2}A }{\sin ^{2} A} }\] \[\frac{ 1-\sin ^{2}A }{sinA }.\frac{ \sin ^{2} A}{ \cos ^{2}A }=\frac{ \cos ^{2}A sinA}{ cosA }=sinAcosA=\frac{ 1 }{2 }\sin(2A)\]
anonymous
  • anonymous
Here is the final answer
anonymous
  • anonymous
So what's the final answer?
anonymous
  • anonymous
\[\frac{ 1 }{ 2 }\sin(2A)\]
anonymous
  • anonymous
does it make sense
anonymous
  • anonymous
uhmm.. yeah sort of. im terrible at this whole simplifying thing. up for a few more?
anonymous
  • anonymous
If you practice, it is not that hard my friend
anonymous
  • anonymous
i got all of the verifying ones but this one..
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anonymous
  • anonymous
I am going for launch. If you stay for 30 minutes i will get back to you.
anonymous
  • anonymous
I'll be here all evening. Thank you! Have a good time :)
anonymous
  • anonymous
\[\frac{ \cos \alpha-\cos \beta }{ \sin \alpha+\sin \beta }+\frac{ \sin \alpha-\sin \beta }{ \cos \alpha+\cos \beta}=\] \[\frac{ \cos ^{2}\alpha-\cos ^{2}\beta+\sin ^{2}\alpha-\sin^2\beta }{(\sin \alpha+\sin \beta)(\cos \alpha +\cos \beta)}\] =\[\frac{ \cos^2 \alpha+\sin^2\alpha-(\sin^2\beta +\cos^2\beta) }{(\sin \alpha+\sin \beta)(\cos \alpha+\cos \beta) }=\frac{ 1-1 }{(\sin \alpha+\sin \beta) (\cos \alpha+\cos \beta) }=0\]
anonymous
  • anonymous
it is verified
anonymous
  • anonymous
how is the denominator verified? nothing was done to it?
anonymous
  • anonymous
since it does not have any significance to simplify the denominator, we can leave it as it is.
anonymous
  • anonymous
How do you know if it has significance to simplify it or not?
anonymous
  • anonymous
In this particular problem, the purpose is to simplify the right hand side to zero. That tells us the numerator of the RHS has to be siplified to zero. our focus should be mostly on working out the numerator.
anonymous
  • anonymous
ohh okay. thanks.
anonymous
  • anonymous
it is my pleasure

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