Simplifying Trigonometric Identities: (sec(x)tan(x))/((tan^2(x))^3/2)

[BigNate]

Hello,

I'm midway through evaluating an integral, and by looking at a solution manual, I see that:

(sec(x)tan(x))/((tan^2(x))^3/2) can simplify to sin^(-2)*cosx. Can someone please explain to me how we can make this jump?

Thanks in advance for your time!

 

[stapel]

I'm midway through evaluating an integral, and by looking at a solution manual, I see that:

(sec(x)tan(x))/((tan^2(x))^3/2) can simplify to sin^(-2)*cosx.

I am guessing that the first expression means this:

. . . . .\(\displaystyle \dfrac{\sec(x)\, \tan(x)}{\left(\tan^2(x)\right)^{3/2}}\)

...and that the second expression means this:

. . . . .\(\displaystyle \sin^{-2}(x)\, \cos(x)\)

What did you get when you simplified the exponential in the denominator, converted everything to sines and cosines, and simplified?

Please show all of your steps. Thank you!

 

[BigNate]

Hello Stapel,

Thank you for your response. With guidance, I got the solution I was looking for:

LHS = (1/cosθ)(sinθ/cosθ) / tan³θ


= (sinθ/cos²θ) / (sin³θ/cos³θ
= sinθ/cos²θ * cos³θ/sin³θ
= cosθ/sin²θ
= cosθsin⁻²θ
= RHS

I am guessing that the first expression means this:

. . . . .\(\displaystyle \dfrac{\sec(x)\, \tan(x)}{\left(\tan^2(x)\right)^{3/2}}\)

...and that the second expression means this:

. . . . .\(\displaystyle \sin^{-2}(x)\, \cos(x)\)

What did you get when you simplified the exponential in the denominator, converted everything to sines and cosines, and simplified?

Please show all of your steps. Thank you!