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cot(theta)-tan(theta) over sin(theta) + cos(theta) equals csc(theta) - sec(theta) how do I proof this?

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Let's use x instead of theta
\[\frac{\cot x-\tan x}{\sin x+\cos x}=\csc x-\sec x\]

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Other answers:

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)
\[(\frac{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}}{\sin x+\cos x})\times \frac{\sin x \cos x}{\sin x \cos x}=\frac{\cos ^2x-\sin ^2x}{\sin x \cos x(\sin x+\cos x)}\]
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
\[\cos ^2x-\sin ^2x=(\cos x-\sin x)(\cos x+\sin x)\]
oh okay that makes sense. so would it be cos-sin(cos+sin)
then that would cancel with the denominator correct?
Yes. sinx + cosx will cancel
which will leave 1/sin(x)+cos(x)
using the reciprocal identities this will equal csc(x)-sec(x) ?
this is where I would use the reciprocal identities?
the cos(x) cancel and the sin(x) cancels?
yay!!!! the cancel and its one,which now you are left with the csc(x) and sec(x) reciprocal identities!! Thank you!! the key idea here that I need to work on was the factoring
thank you !:)
thanks ! :)

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