How do you prove? #[cos(pi/2-x)-2cos(pi/2-x)sin^2x+cos(pi/2-x)sin^4x]sec^5(-x) = tanx#

#(cos(pi/2-x)-2cos(pi/2-x)sin^2x+cos(pi/2-x)sin^4x)sec^5(-x)=tanx#

Let's start by applying the cos difference identity:
#Cos(alpha-theta)=cosalphacostheta+sinalphasintheta#
#(cos(pi/2)cosx+sin(pi/2)sinx)-2(cos(pi/2)cosx+sin(pi/2)sinx)sin^2x+(cos(pi/2)cosx+sin(pi/2)sinx)sin^4x)sec^5(-x)=tanx#

Let's simplify:
#(cancel(0*cosx)+1*sinx)-2(cancel(0*cosx)+1*sinx)sin^2x+(cancel(0*cosx)+1*sinx)sin^4x)sec^5(-x)=tanx#

#(sinx)-2(sinx)sin^2x+(sinx)sin^4x)sec^5(-x)=tanx#

Note that secant is in fact an even function:
#(sinx-2sin^3x+sin^5x)sec^5(x)= tanx#

Let's try factoring by GCF:
#sinx(1-2sin^2x+sin^4x)sec^5(x)= tanx#

We can now factor the parenthesis:
#sinx(1-sin^2x)(1-sin^2x)sec^5(x)= tanx#

A very familiar modified Pythagorean identity will now be implemented:
#1-sin^2x=cos^2x#

Substitute:
#sinx(cos^2x)(cos^2x)sec^5(x)= tanx#

Simplify:
#sinx*cos^4x*sec^5x= tanx#

Apply the secant reciprocal identity:
#1/cosx= secx#

Substitute:
#sinx*cos^4x*1/cos^5x= tanx#

Simplify:
#sinx*cancel(cos^4x)*1/cos^cancel(5)x= tanx#

#sinx*1/cosx=tanx#

#sinx/cosx=tanx#

#tanx=tanx#

The expression

#[cos(pi/2-x)-2cos(pi/2-x)sin^2x+cos(pi/2-x)sin^4x]sec^5(-x) #

simplifies considerably if you use #cos(pi/2-x) = sin x# and #sec(-x) = sec(x)#, becoming

#(sinx-2sin^3x+sin^5x)sec^5x #
#= sin x(1-2sin^2x+sin^4x)sec^5x #
#=sinx(1-sin^2x)^2sec^5x#
#=sinx cos^4xsec^5x = sinxsecx=tanx#