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Precalculus 15 Online
OpenStudy (anonymous):

Verify the identity. cot (x-pi/2)= -tan x PLEASE HELP

OpenStudy (anonymous):

change cot to sin and cos

OpenStudy (anonymous):

@jonask how do you do hat ? I need help with this

OpenStudy (anonymous):

\[\cot \theta=\frac{ \cos \theta }{ \sin \theta }\] use the theta above x-pi/2

OpenStudy (anonymous):

So cot theta = x-pi/2?

OpenStudy (anonymous):

no cot theta=cot (x-pi/2)

OpenStudy (anonymous):

now what ?

OpenStudy (anonymous):

use the identity above that writes cot as cos /sin

OpenStudy (anonymous):

I thought the result would be cot theta = -tanx

OpenStudy (anonymous):

lets use this id first\[\cot(x-\pi/2)=\frac{\cos (x-\pi/2)}{\sin (x-\pi/2)}\]

OpenStudy (anonymous):

-tanx= cot(x-pi/2)

OpenStudy (anonymous):

right ? im not sure of the steps

OpenStudy (anonymous):

@jonask

OpenStudy (anonymous):

we are proving it here so according to the last step we simplify as\[\frac{ -\sin x }{ \cos x }=-\tan x\] this equal to -tan x

OpenStudy (anonymous):

So what do I do for the steps i need help working it out ... @jonask

OpenStudy (anonymous):

\[\cot (x-\pi/2)=\frac{ \cos (x-\pi/2) }{ \sin(x-\pi/2) }=\frac{ \cos(\pi/2-x )}{ -\sin(\pi/2-x) }=\frac{ \sin x }{ -\cos x }=-\tan x\]

OpenStudy (anonymous):

THANK YOU SO MUCH !!!! @Jonask

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