Diana0920
cot(theta)-tan(theta) over sin(theta) + cos(theta) equals csc(theta) - sec(theta)
how do I proof this?
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Mertsj
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Let's use x instead of theta
Diana0920
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okay
Mertsj
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\[\frac{\cot x-\tan x}{\sin x+\cos x}=\csc x-\sec x\]
Mertsj
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Do I have the problem stated correctly?
Diana0920
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yes :)
Diana0920
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would I have to replace cot(x) with cos(x)/sin(x)
Diana0920
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and tan(x) with sin(x)/cos(x)
Mertsj
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\[(\frac{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}}{\sin x+\cos x})\times \frac{\sin x \cos x}{\sin x \cos x}=\frac{\cos ^2x-\sin ^2x}{\sin x \cos x(\sin x+\cos x)}\]
Mertsj
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Can you get it from there?
Diana0920
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sadly no?
can you explain how I can simplify it more please?
Mertsj
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Factor the numerator and then the cosx + sinx will cancel.
alejandrop95
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I'm still lost :(
Mertsj
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@alejandrop95 This is not your question
Diana0920
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how would I factor it though? I'm sorry I am new to trig.
Mertsj
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Can you factor this:
\[a^2-b^2\]
Diana0920
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I don't think you can because it doesn't have the same coefficient
Mertsj
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\[a^2−b^2=(a-b)(a+b)\]
Mertsj
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\[\cos ^2x-\sin ^2x=(\cos x-\sin x)(\cos x+\sin x)\]
Diana0920
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oh okay that makes sense. so would it be cos-sin(cos+sin)
Mertsj
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yes
Diana0920
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then that would cancel with the denominator correct?
Mertsj
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Yes. sinx + cosx will cancel
Diana0920
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which will leave 1/sin(x)+cos(x)
Mertsj
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no
Diana0920
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using the reciprocal identities this will equal csc(x)-sec(x) ?
Mertsj
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|dw:1382579245755:dw|
Diana0920
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cos(x)-sin(x)/sin(x)+cos(x0
Diana0920
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this is where I would use the reciprocal identities?
Mertsj
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|dw:1382579370642:dw|
Diana0920
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the cos(x) cancel and the sin(x) cancels?
Mertsj
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|dw:1382579509266:dw|
Diana0920
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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
Diana0920
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thank you !:)
alejandrop95
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thanks !
:)
Mertsj
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yw