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How can I prove that : (cosx +1)/sinx = (sinx-cosx-1)/ (1-cosx-sinx)

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xetabeta | Student, Kindergarten | Salutatorian

Posted March 11, 2011 at 2:51 PM via web

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How can I prove that : (cosx +1)/sinx = (sinx-cosx-1)/ (1-cosx-sinx)

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted March 11, 2011 at 3:01 PM (Answer #1)

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We have to prove that (cos x +1)/sin x = (sin x - cos x - 1)/ (1- cos x - sin x)

Let us assume the identity is true:

(cos x +1)/sin x = (sin x - cos x - 1)/ (1- cos x - sin x)

=> (cos x +1)(1- cos x - sin x) = (sin x - cos x - 1)*(sin x)

=> 1- cos x - sin x + cos x - (cos x)^2 - (cos x)*(sin x) = (sin x)^2 - sin x * cos x - sin x

Eliminate the common terms

=> 1 - (cos x)^2 = (sin x)^2

=> 1 = (sin x)^2 + (cos x)^2

=> 1 = 1

This shows our initial assumption was correct and the given identity (cos x +1)/sin x = (sin x - cos x - 1)/ (1- cos x - sin x) is true for all values of x.

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giorgiana1976 | College Teacher | Valedictorian

Posted March 11, 2011 at 3:24 PM (Answer #2)

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We'll cross multiply the terms of the fractions:

sin x*(sin x - cos x - 1) = (cosx +1)(1 - cos x - sin x)

We'll remove the brackets:

(sin x)^2 - sin x*cos x - sin x = cos x - (cos x)^2 - sin x*cos x + 1 - cos x - sin x

We'll eliminate sin x*cos x both sides:

(sin x)^2  - sin x = cos x - (cos x)^2 + 1 - cos x - sin x

We'll eliminate like terms both sides:

 (sin x)^2 = - (cos x)^2 + 1

We'll add (cos x)^2 both sides:

 (sin x)^2 + (cos x)^2 = 1

From the Pythagorean identity, we know that (sin x)^2 + (cos x)^2 = 1.

The given identity is true: (cosx +1)/sinx = (sinx-cosx-1)/ (1-cosx-sinx).

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