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Precalculus 14 Online
OpenStudy (anonymous):

Write the trigonometric expression in terms of sine and cosine, and then simplify. csc2 x − cot2 x

hartnn (hartnn):

whats csc x in terms of sin x ?

OpenStudy (anonymous):

thats what i do not know

hartnn (hartnn):

csc x = 1/sin x

hartnn (hartnn):

and cot x in terms of sin and cos is ?

OpenStudy (anonymous):

cos/sin

OpenStudy (anonymous):

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

hartnn (hartnn):

that is correct, now just put both the formulas in the question, csc x = 1/ sin x and cot x = cos x / sin x

hartnn (hartnn):

or u can see @Zekarias solution and ask if u don't understand any of thhe step...

OpenStudy (anonymous):

so 1-cos^2x = sin ^2 x ?

OpenStudy (anonymous):

yep

hartnn (hartnn):

thats from the identity \(\huge sin^2x+cos^2x = 1\)

OpenStudy (anonymous):

ohh that makes sense. i remember now thanks

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