How do you verify #(1+tanx)/(1-tanx) = (cosx+sinx)/(cosx-sinx)#?

1 Answer
Mar 9, 2016

see explanation

Explanation:

manipulate the left side

#rArr (1+tanx)/(1-tanx) = (1+sinx/cosx)/(1-sinx/cosx)#

rewrite numerator / denominator as single fractions.

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

multiply the fractions by ' inverting' the fraction on denominator.

#rArr (cosx+sinx)/cancel(cosx) xx (cancel(cosx))/(cosx-sinx)#

#= (cosx+sinx)/(cosx-sinx) = " right side " #