How do you prove the identity #(1-sinx)/cosx=cosx/(1+sinx)#?

Question

65801 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-15T12:32:42

By multiplying both numerator and denominator by #1+sinx # and using the difference of squares the result follows quickly.

multiply the LHS , top and bottom by #(1+sinx)#

#((1-sinx)(1+sinx))/(cosx(1+sinx))#

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

but #sin^2x+cos^x=1#

#:. =(cos^2x)/(cosx(1+sinx))#

# =(cancel(cosx)(cosx))/(cancelcosx(1+sinx))#

as required.