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prove the identity sin (angle)^2/ 1-cos(angle) =sec(angle)+1/sec(angle) 

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ssdixon | Student, Undergraduate | Salutatorian

Posted May 8, 2012 at 3:59 AM via web

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prove the identity sin (angle)^2/ 1-cos(angle) =sec(angle)+1/sec(angle) 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 8, 2012 at 5:33 AM (Answer #1)

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You need to prove that the given expression is identity, hence, you should live the left side and you should focus on the right side such that:

`(sin^2 alpha)/(1 - cos alpha) = (sec alpha + 1)/sec alpha`

Notice that not the angle is raised to square, but the sine function and notice the letter `alpha`  instead of angle.

You need to remember that `sec alpha = 1/cos alpha` .

`(sin^2 alpha)/(1 - cos alpha) = (1/(cos alpha) + 1)/(1/cos alpha )`

`(sin^2 alpha)/(1 - cos alpha) = (1+cos alpha)`

You need to multiply both sides by `1 - cos alpha`  such that:

`sin^2 alpha = (1 - cos alpha)(1 + cos alpha)`

You need to convert the special product `(1 - cos alpha)(1 + cos alpha)`  in a difference of squares such that:

`sin^2 alpha = 1 - cos^2 alpha`

You need to add `cos^2 alpha ` both sides such that:

`sin^2 alpha + cos^2 alpha = 1`

Hence, the last line, `sin^2 alpha + cos^2 alpha = 1,`   proves that given expression represents an identity.

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malagala | College Teacher | (Level 2) Adjunct Educator

Posted May 8, 2012 at 4:51 AM (Answer #2)

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sin(angle)^2 / 1-cos(angle)

as sin(angle)^2 + cos(angle)^2 = 1

we can rewrite as,

=>{1 - cos(angle)^2 }/ 1-cos(angle)

=> (1-cos(angle))(1+cos(angle)) / (1 - cos(angle))

=> 1 + cos(angle)

as cos(angle) = 1/sec(angle)

=> 1 + 1/sec(angle)

=> (sec(angle) + 1) / sec(angle)

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