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Can someone please help me establish this identity?
[tex]\cos \theta (\tan \theta + \cot \theta) = \csc \theta[/tex]
[tex]\cos \theta (\tan \theta + \cot \theta) = \csc \theta[/tex]
The easy ones always get me :\irony of truth said:So, are you proving this identity?
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
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.TD said:How did you end up with that?
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