Question

Question #f68d0

Answer 1

`secx-sinxtanx`

Explanation:

Since `cscx=1/sinx` and `cotx=cosx/sinx`, substitute these in to get `(1/sinx - sinx)/(cosx/sinx)`

Multiply the reciprocal of the denominator.

`1/sinx - sinx*sinx/cosx`

We end up with this after using the distributive property.

`sinx/(sinxcosx) - sin^2x/cosx`

Cancel out the two `sinx`s on the left.

`1/cosx-sin^2x/cosx`

Since `1/cosx=secx` and `sinx/cosx=tanx`, the simplified answer is

`secx-sinxtanx`

EDIT: Ignore this, not fully simplified. See Scott's answer.

Answer 2

`cos x`

Explanation:

First put everything in terms of sine and cosine

`csc x = 1/sin x` and `cot x = cos x/sin x`, so `(csc x - sin x)/cot x = (1/sin x - sin x)/(cos x/sin x) = (1/sin x - sin x)(sin x/cos x)`

Then distribute multiplication over subtraction

`(1/sin x - sin x)(sin x/cos x) = (1/sin x * sin x/cos x) - (sin x * sin x/cos x)`

Multiply inside the parentheses

`(1/sin x * sin x/cos x) - (sin x * sin x/cos x) = (sin x /(sin x cos x)) - ((sin x sin x)/cos x)`

And simplify

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

Using the Pythagorean identity `sin^2x + cos^2x = 1`, we know that `1 - sin^2x = cos^2x`, so substitute that in

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

And finally, simplify

`cos^2x/cos x = cos x/1 = cos x`