anonymous
  • anonymous
Identities- Ugh! 1÷cscθ−cotθ=1+cosθ÷sinθ
Trigonometry
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anonymous
  • anonymous
Identities- Ugh! 1÷cscθ−cotθ=1+cosθ÷sinθ
Trigonometry
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

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anonymous
  • anonymous
r u missing any brackets?
ZeHanz
  • ZeHanz
It should be:\[\frac{ 1 }{ \csc \theta - \cot \theta }=\frac{ 1+\cos \theta }{ \sin \theta }\]Let's begin at the left hand side:\[\frac{ 1 }{ \csc \theta - \cot \theta }=\frac{ 1 }{ \frac{ 1 }{ \sin \theta }-\frac{ \cos \theta }{ \sin \theta } }=\frac{ 1 }{ \frac{ 1-\cos \theta }{ \sin \theta } }=\frac{ \sin \theta }{ 1- \cos \theta }\]This looks much better, although it is not quite the right hand side yet... Hint: multiply numerator and denominator of the last fraction with (1+cosθ)/(1+cosθ):\[\frac{ \sin θ }{ 1-\cos θ } \cdot \frac{ 1+\cos θ }{ 1+\cos θ }= ...\]After just a few small steps, you'll have got to the right hand side of the identity!

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