Question

How do you show the `secx + 1 + (1 - tan^2x)/(secx -1) = cosx/(1 - cosx)`?

Answer 1

We have:

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

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

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

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

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

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

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

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

`cosx/(1 - cosx) =cosx(1 - cosx)`

`LHS = RHS`

Hopefully this helps!

Answer 2

To prove the identity, we'll need these identities:

`secx=1/cosx`

`sec^2x=tan^2x+1`

I'll start with the left side of the equation and manipulate it until it equals the right side:

`LHS=secx + 1 + (1 - tan^2x)/(secx -1)`

`color(white)(LHS)=((secx + 1)(secx-1))/(secx-1) + (1 - tan^2x)/(secx -1)`

`color(white)(LHS)=(color(blue)(sec^2x)-1)/(secx-1) + (1 - tan^2x)/(secx -1)`

`color(white)(LHS)=(color(blue)(tan^2x+1)-1)/(secx-1) + (1 - tan^2x)/(secx -1)`

`color(white)(LHS)=(tan^2xcolor(red)cancelcolor(black)(color(black)+1-1))/(secx-1) + (1 - tan^2x)/(secx -1)`

`color(white)(LHS)=(tan^2x)/(secx-1) + (1 - tan^2x)/(secx -1)`

`color(white)(LHS)=(tan^2x+1 - tan^2x)/(secx -1)`

`color(white)(LHS)=(color(red)cancelcolor(black)(tan^2x)+1 color(red)cancelcolor(black)(color(black)- tan^2x))/(secx -1)`

`color(white)(LHS)=1/(color(blue)secx-1)`

`color(white)(LHS)=1/(color(blue)(1/cosx)-1)`

`color(white)(LHS)=1/((1/cosx-1))color(red)(*cosx/cosx)`

`color(white)(LHS)=cosx/(1-cosx)`

`color(white)(LHS)=RHS`

That's the proof. Hope this helped!