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

2 Answers
Noah G
Apr 7, 2018

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!

Aviv S.
Apr 7, 2018

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!

Related questions
See all questions in Proving Identities
Impact of this question
3041 views around the world
You can reuse this answer
Creative Commons License