Prove that #(tanx-cotx)/(tanx+cotx)=sin^2x-cos^2x#?

1 Answer
Nov 26, 2016

Please see below.

Explanation:

#(tanx-cotx)/(tanx+cotx)#

= #(sinx/cosx-cosx/sinx)/(sinx/cosx+cosx/sinx)# - (converting tan ad cot ratios into sin and cos ratios)

= #((sin^2x-cos^2x)/(sinxcosx))/((sin^2x+cos^2x)/(sinxcosx))# - (adding as in fractions)

= #(sin^2x-cos^2x)/(sinxcosx)xx(sinxcosx)/((sin^2x+cos^2x)#

= #(sin^2x-cos^2x)/cancel(sinxcosx)xxcancel(sinxcosx)/1# - (as #sin^2x+cos^2x=1# and cancelling like terms in numerator and denominator)

= #sin^2x-cos^2x#