- #1

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[tex]\cos \theta (\tan \theta + \cot \theta) = \csc \theta[/tex]

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- Thread starter cscott
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- #1

- 782

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[tex]\cos \theta (\tan \theta + \cot \theta) = \csc \theta[/tex]

- #2

- 90

- 0

Express your tangent and cotangent in terms of sine and cosine. Get their LCD... and your numerator becomes a well-known trigonometric identity..

Can you continue from here? :D

- #3

- 782

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irony of truth said:

Express your tangent and cotangent in terms of sine and cosine. Get their LCD... and your numerator becomes a well-known trigonometric identity..

Can you continue from here? :D

The easy ones always get me :\

Thanks!

- #4

- 782

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[tex]\frac{1 + \tan \theta}{1 - \tan \theta} = \frac{\cot \theta + 1}{\cot \theta - 1}[/tex]

I'm so bad at proofs

- #5

mezarashi

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- #6

TD

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Try, if you get stuck, show us!

- #7

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[tex]\frac{\cos^2 \theta + \sin \theta \cos \theta}{\cos^2 \theta - \sin \theta \cos \theta}[/tex]

or

[tex]\frac{\cot^2 \theta + \cot \theta}{\cot^2 \theta - \cot \theta}[/tex]

How do I continue?

- #8

TD

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For the LHS:

[tex]\frac{{1 + \tan \theta }}{{1 - \tan \theta }} = \frac{{1 + \frac{{\sin \theta }}{{\cos \theta }}}}{{1 - \frac{{\sin \theta }}{{\cos \theta }}}} = \frac{{\frac{{\cos \theta + \sin \theta }}{{\cos \theta }}}}{{\frac{{\cos \theta - \sin \theta }}{{\cos \theta }}}} = \frac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}[/tex]

Now try the RHS

- #9

- 782

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TD said:

For the LHS:

[tex]\frac{{1 + \tan \theta }}{{1 - \tan \theta }} = \frac{{1 + \frac{{\sin \theta }}{{\cos \theta }}}}{{1 - \frac{{\sin \theta }}{{\cos \theta }}}} = \frac{{\frac{{\cos \theta + \sin \theta }}{{\cos \theta }}}}{{\frac{{\cos \theta - \sin \theta }}{{\cos \theta }}}} = \frac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}[/tex]

Now try the RHS

Silly me - I just multiplied out the numerator by the reciprocal of the denomenator instead of just canceling out the cosines. If you factor the top and bottom of my expression you end up with what your answer. If I do this using 1/cot = tan I end up with the RHS.

Don't I need to continue with the LHS until I get the right or vice versa?

- #10

TD

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- #11

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Ah, I see. Thank you both of you.

- #12

TD

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No problem

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