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 " #