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# need this answered really quick need it for tomorrow

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sec^6(x)[sec(x)tan(x)] -sec^4(x)[sec(x)tan(x)] = sec^5(x)tan^3(x)

thanks.

Guest Jun 4, 2017
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sec6(x) sec(x) tan(x)  -  sec4(x) sec(x) tan(x)

=  sec7(x) tan(x)   -   sec5(x) tan(x)               Factor out   sec5(x) tan(x)   .

=  sec5(x) tan(x) * ( sec2(x)  -  1 )                 Rewrite secant as 1/cosine

= sec5(x) tan(x) * ( 1/cos2(x)  -  1 )               Get a common denominator and combine the fractions.

= sec5(x) tan(x) * ( 1/cos2(x)  -  cos2(x)/cos2(x) )

= sec5(x) tan(x) * ( [ 1 -  cos2(x) ] /cos2(x) )    Rewite   1 - cos2(x)   as   sin2(x)

= sec5(x) tan(x) * ( sin2(x) /cos2(x) )

= sec5(x) tan(x) * tan2(x)

= sec5(x) tan3(x)

hectictar  Jun 4, 2017

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