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# Calculus/Trignometry help

Question
Hi,

I can only do work on one side to prove that its end result is equal to the other side.

Thanks a lot.

(cscx-cotx)^2   = ( 1/sinx    - cos x / sin x ) ^2    (change every function to sin and cosine)
=  ((1-cos x )/sin x   )^2          (same denominator, so we can add numerators)
= (1-cos x ) ^ 2/   (sin x)^2          (power distributes over division and multiplication )
= (1-cos x ) ^ 2/   ( 1 - (cos x)^2  )  (some trig identity that says sin^2 A + cos^2 A = 1)
= (1-cos x ) ^ 2/   ( 1 - cos x)  (1+ cos x)   (factor the denominator since it's
the difference of two squares.)

I leave it to you now to cancel simplify and finish it up.  Happy holidays!

Calculus

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#### Amos

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I can answer all calculus question. I am a Math Lecturer and I teach Math in a College, usually Calculus 1, Calculus 2 and Calculus 3 and Linear Algebra.

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