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Question
Prove the following:
`(sin^3(pi + x)sec^2(pi - x)tan(2pi - x))/(cos^2(pi/2 + x)sin(pi - x)"cosec"^2 - x)` = tan3x
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Solution
L.H.S. = `(sin^3(pi + x)sec^2(pi - x)tan(2pi - x))/(cos^2(pi/2 + x)sin(pi - x)"cosec"^2 (- x))`
= `([sin(pi + x)]^3 [sec(pi - x)]^2 tan(2pi - x))/([cos(pi/2 + x)]^2 sin(pi - x)*(-"cosec")^2`
= `((-sinx)^3(-secx)^2(-tanx))/((-sinx)^2*sinx*"cosec"^2x)`
= `((-sin^3x)*sec^2x*(-tanx))/(sin^2x*sinx*1/sin^2x)`
= `(sin^3x*sec^2x*tanx)/sinx`
= `sin^2x*1/cos^2x tanx`
= tan2x.tanx
= tan3x
= R.H.S.
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