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

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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.

malagala | College Teacher | (Level 2) Adjunct Educator

Posted on

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)