Question #f68d0

2 Answers
Mar 23, 2017

Answer:

#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.

Mar 23, 2017

Answer:

#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#