Question #f68d0

Question

2744 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-23T23:11:13

$secx-sinxtanx$

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 · 2017-03-23T23:23:08

$cos x$

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

$cos^2x/cos x = cos x/1 = cos x$