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Let's use x instead of theta

okay

\[\frac{\cot x-\tan x}{\sin x+\cos x}=\csc x-\sec x\]

Do I have the problem stated correctly?

yes :)

would I have to replace cot(x) with cos(x)/sin(x)

and tan(x) with sin(x)/cos(x)

Can you get it from there?

sadly no?
can you explain how I can simplify it more please?

Factor the numerator and then the cosx + sinx will cancel.

I'm still lost :(

@alejandrop95 This is not your question

how would I factor it though? I'm sorry I am new to trig.

Can you factor this:
\[a^2-b^2\]

I don't think you can because it doesn't have the same coefficient

\[a^2âˆ’b^2=(a-b)(a+b)\]

\[\cos ^2x-\sin ^2x=(\cos x-\sin x)(\cos x+\sin x)\]

oh okay that makes sense. so would it be cos-sin(cos+sin)

yes

then that would cancel with the denominator correct?

Yes. sinx + cosx will cancel

which will leave 1/sin(x)+cos(x)

no

using the reciprocal identities this will equal csc(x)-sec(x) ?

|dw:1382579245755:dw|

cos(x)-sin(x)/sin(x)+cos(x0

this is where I would use the reciprocal identities?

|dw:1382579370642:dw|

the cos(x) cancel and the sin(x) cancels?

|dw:1382579509266:dw|

thank you !:)

thanks !
:)

yw