Question

Prove it: `tan^5x=((1/(1-sinx)^2)-(1/(1+sinx)^2))/((1/(1-cosx)^2)-(1/(1+cosx)^2)`?

Answer 1

To prove

`tg^5x=((1/(1-sinx)^2)-(1/(1+sinx)^2))/((1/(1-cosx)^2)-(1/(1+cosx)^2)`

RHS

`=((1/(1-sinx)^2)-(1/(1+sinx)^2))/((1/(1-cosx)^2)-(1/(1+cosx)^2)`

`=(((1+sinx)^2-(1-sinx)^2)/(1-sin^2x)^2)/(((1+cosx^2)-(1-cosx)^2)/(1-cos^2x)^2)`

`=((4sinx)/cos^4x)/((4cosx)/(sin^4x))`

`=sin^5x/cos^5x=tan^5x=LHS`

Proved

Answer 2

This is one of those proofs that is easier to work from right to left. Start with:

`((1/(1-sinx)^2)-(1/(1+sinx)^2))/((1/(1-cosx)^2)-(1/(1+cosx)^2)`

Multiply numerator and denominator of the embedded fractions by the "conjugates" (e.g. `1pmsinx` on `1 ∓ sinx`). You get that, e.g., `(1+sinx)(1-sinx) = 1-sin^2x`.

`= (((1+sinx)/((1-sin^2x)(1-sinx)))-((1-sinx)/((1-sin^2x)(1+sinx))))/(((1+cosx)/((1-cos^2x)(1-cosx)))-((1-cosx)/((1-cos^2x)(1+cosx)))`

Repeat the previous step to simplify the denominator in the embedded fractions further:

`= (((1+sinx)^2/((1-sin^2x)^2))-((1-sinx)^2/((1-sin^2x)^2)))/(((1+cosx)^2/((1-cos^2x)^2))-((1-cosx)^2/((1-cos^2x)^2))`

Use the identities `1-sin^2x = cos^2x` and `1-cos^2x = sin^2x` to get:

`= (((1+sinx)^2/(cos^4x))-((1-sinx)^2/(cos^4x)))/(((1+cosx)^2/(sin^4x))-((1-cosx)^2/(sin^4x))`

Combine fractions and flip to multiply the reciprocals:

`= (((1+sinx)^2-(1-sinx)^2)/(cos^4x))/(((1+cosx)^2-(1-cosx)^2)/(sin^4x))`

`= ((1+sinx)^2-(1-sinx)^2)/(cos^4x)*(sin^4x)/((1+cosx)^2-(1-cosx)^2)`

Expand the squared terms:

`= (cancel(1)+2sinx+cancel(sin^2x)-(cancel(1)-2sinx+cancel(sin^2x)))/(cos^4x)*(sin^4x)/(cancel(1)+2cosx+cancel(cos^2x)-(cancel(1)-2cosx+cancel(cos^2x)))`

`= (cancel(4)sinx)/(cos^4x)*(sin^4x)/(cancel(4)cosx)`

`= color(blue)(tan^5x)`