Question

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

Thank you!

Answer 1

Verified below

Explanation:

`(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`

Answer 2

See below

Explanation:

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`