anonymous
  • anonymous
Prove this identity sin^2 Θcsc^2 Θ = sin^2 Θ = cos^2 Θ
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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myininaya
  • myininaya
Think you have too many equality symbols
anonymous
  • anonymous
That is what the question I got for homework is asking. I don't understand it.
Mertsj
  • Mertsj
You copied it wrong. It should be: \[\sin ^2x \csc ^2x-\sin^2x=\cos^2x\]

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myininaya
  • myininaya
\[\sin^2(\theta)+\cos^2(\theta)=1\] \[1+\cot^2(\theta)=\csc^2(x) => \csc^2(x)-1=\cot^2(x)\] So maybe it is suppose to say \[\sin^2(\theta) \csc^2(\theta)-\sin^2(\theta)=\cos^2(\theta)\] \[\sin^2(\theta)(\csc^2(\theta)-1)=\sin^2(\theta) \cdot \cot^2(x)=\sin^2(\theta) \cdot \frac{\cos^2(\theta)}{\sin^2(\theta)}\]
anonymous
  • anonymous
No this is the problem: sin^2 Θcsc^2 Θ = sin^2 Θ = cos^2 Oh well, I am going to ask my teacher if she wrote the equation wrong
Mertsj
  • Mertsj
\[\sin ^2\theta \times\frac{1}{\sin ^2\theta}- \sin ^2\theta=\cos ^2\theta\]
Mertsj
  • Mertsj
She did.
myininaya
  • myininaya
she totally wrote it wrong so i totally agree with mr. mert here she meant to write the way we said just change it you might be only one who does the problem (And might get bonus :))
anonymous
  • anonymous
Okay I will do that Thank You:)

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