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Mathematics 21 Online
OpenStudy (anonymous):

last one! need help verifying 1 - tan cos/ csc = cos^2

OpenStudy (anonymous):

\[1-\frac{ \tan(x)\cos(x) }{ \csc(x) }=1-\frac{ \frac{ \sin(x) }{ \cos(x) }\cos(x) }{ \frac{ 1 }{ \sin(x) } }\]\[=1-\sin^{2}x=\cos^{2}x\]

OpenStudy (anonymous):

you need some parenthesis

OpenStudy (anonymous):

@Zekarias can you explain what you did?

OpenStudy (anonymous):

wright tan(x) and csc(x) as sin(x)/cos(x) and 1/sin(x), first. Can u do that?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Then cancel cos(x) with cos (x). What will u have?

OpenStudy (anonymous):

you'll have 1 - sin divided by 1/sin

OpenStudy (anonymous):

what is sin(x)/1/sin(x)?

OpenStudy (anonymous):

sin^2

OpenStudy (anonymous):

1-sin^2(x) = cos ^2(x) wright?

OpenStudy (anonymous):

oh i understand now! its on eof the identities!

OpenStudy (anonymous):

thank you!

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