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

Answer 1

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.