`cot^3(t)/csc(t) = cos(t)(csc^2(t) - 1)` Verfiy the identity.

Textbook Question

Chapter 5, 5.2 - Problem 18 - Precalculus (3rd Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember that `1/(csc t) = sin t` , hence, replacing `sin t` for `1/(csc t)` to the left side, yields:

`sin t*cot^3 t = (cos t)*(csc^2 t - 1)`

You need to replace `(cos^3 t)/(sin^3 t) ` for `cot^3 t` , to the left side:

`sin t*(cos^3 t)/(sin^3 t) = (cos t)*(csc^2 t - 1)`

Reducing by sin t to the left, yields:

`(cos^3 t)/(sin^2 t) = (cos t)*(csc^2 t - 1)`

Reducing by cos t both sides, yields:

`(cos^2 t)/(sin^2 t) = (csc^2 t - 1)`

`cot^2 t = csc^2 t - 1`

Replacing `1/(sin^2 t)` for` csc^2 t` yields:

`cot^2 t = 1/(sin^2 t) - 1 => cot^2 t + 1 = 1/(sin^2 t)`

Notice that the identity `cot^2 t + 1 = 1/(sin^2 t)` is valid, since it is derived from `sin^2 t + cos^2 t = 1` .

Hence, verifying if the given identity` (cot^3 t)/(csc t) = (cos t)*(csc^2 t - 1)` is valid, yields that it is.

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scisser | (Level 3) Honors

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`cot = cos/sin, so cot^3 = cos^3 / sin ^3 `

`csc = 1/sin so 1/csc = sin `

Left side:

`cot^3t/csct `

`=(cos^3t / sin^3t) * sint `

`=cos^3t / sin^2t`

Right side:

`1+cot^2 = csc^2 `

`csc^2 -1 = cot^2 `

so, right side:

`cost(csc^2t-1) `

`=cost(cot^2t) `

`=cost( cos^2t/sin^2t) `

`=cos^3t/sin^2t `

Thus, both sides are the same

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