bendent1234
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#1
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I have a homework question where I have to fully simplify:

 \frac {a \mathrm{cosec} \theta} {(a \mathrm{cosec} \theta)^2 - a^2}

I have simplified this to:

 \frac {\mathrm{cosec} \theta} {a \cot^2 \theta}

The answer given in the book is:

 \frac {\tan \theta \sec \theta} {a}

Can anyone help with the steps?
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DFranklin
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From your \dfrac{\cosec \theta}{a \cot^2 \theta} expression I'd just rewrite everything in term so sin/cos and cancel down. It should be easy to go back to the tan t sec t form at the tend.
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Darth_Narwhale
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yeah. Also, when you have the inverse trig functions such as cot on the bottom, you should know to simplify by putting the normal function on top.
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RDKGames
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(Original post by Darth_Narwhale)
yeah. Also, when you have the inverse trig functions such as cot on the bottom, you should know to simplify by putting the normal function on top.
\cot(x) isn't an inverse trigonometric function.
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Darth_Narwhale
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(Original post by RDKGames)
\cot(x) isn't an inverse trigonometric function.
Yeah I know, I thought someone would pick me up on that . Is reciprocal function a better term?
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RDKGames
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(Original post by Darth_Narwhale)
Yeah I know, I thought someone would pick me up on that . Is reciprocal function a better term?
Yep
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bendent1234
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I keep ending up with:

 \frac {\sin \theta \sec^2 \theta} {a}

Which is not the answer. Anyone got any advice?
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RDKGames
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(Original post by bendent1234)
I keep ending up with:

 \frac {\sin \theta \sec^2 \theta} {a}

Which is not the answer. Anyone got any advice?
\sin(x) \sec(x) = \tan(x)
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bendent1234
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(Original post by RDKGames)
\sin(x) \sec(x) = \tan(x)

Oh of course it is
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