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Is the following an identity: csc x + cot x = sin x/1-cos x

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paul1234 | Student | eNoter

Posted April 11, 2011 at 2:10 PM via web

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Is the following an identity: csc x + cot x = sin x/1-cos x

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted April 11, 2011 at 2:18 PM (Answer #1)

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To prove csc x + cot x = sin x /(1 - cos x), let's start from the left hand side

csc x + cot x

we know that csc x = 1/ sin x and cot x = (cos x/sin x)

=> 1/ sin x + cos x / sin x

=> (1 + cos x)/sin x

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

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

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

=> sin x/(1 - cos x)

which is the right hand side.

This is valid for all value of x except where we get indeterminate forms like 1/0 when sin x = 0 or at x = k*pi

This proves that csc x + cot x = sin x /(1 - cos x). It is valid for all x except x = k*pi

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giorgiana1976 | College Teacher | Valedictorian

Posted April 11, 2011 at 5:56 PM (Answer #2)

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We'll re-write the expression using brackets:

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

Now, we'll multiply both sides by (1-cos x):

(1 - cos x)(csc x + cot x) = sin x

We'll remove the brackets using FOIL method:

cscx + cot x - cos x* csc x - cos x*cot x = sin x

We'll substitute csc x = 1/sin x:

1/sin x + cot x - cos x/sin x - cos x*cot x = sin x

1/sin x + cot x - cot x - cos x*cot x = sin x

We'll eliminate like terms:

1/sin x - cos x*cot x = sin x

But cot x = cos x/sin x

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

Since the fractions have the same denominator, we'll re-write the left side:

[1 -(cos x)^2]/sin x = sin x

But, from Pythagorean identity, we'll get:

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

The identity will become:

(sin x)^2/sin x = sin x

W'll simplify and we'll get:

sin x = sin x

Since both sides are equal, then the identity csc x + cot x = sin x/1-cos x is verified, for any real value of x.

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