How do you simplify #(1- tan^2θ) /( 1+ tan^2 θ)#?

2 Answers
Mar 25, 2018

#2cos^2theta-1# or #cos2theta#

Explanation:

Using the identities:
#1+tan^2theta= sec^2theta#
#1/sectheta= costheta#
#tantheta= sintheta/costheta#
#sin^2theta= 1-cos^2theta#
#2cos^2theta-1= cos2theta#

Start:
#(1-tan^2theta)/(1+tan^2theta)=#

#(1-tan^2theta)/(sec^2theta)=#

Split the numerator:

#1/sec^2theta-tan^2theta/sec^2theta=#

#cos^2theta-sin^2theta/cancel(cos^2theta)*cancel(cos^2theta)=#

#cos^2theta-(1-cos^2theta)=#

#2cos^2theta-1=#

#cos2theta#

Mar 25, 2018

#cos2theta#

Explanation:

#"using the "color(blue)"trigonometric identities"#

#•color(white)(x)tantheta=sintheta/costheta#

#•color(white)(x)sin^2theta+cos^2theta=1#

#•color(white)(x)cos^2theta-sin^2theta=cos2theta#

#rArr(1-tan^2theta)/(1+tan^2theta)#

#=(1-sin^2theta/cos^2theta)/(1+sin^2theta/cos^2theta)xxcos^2theta/cos^2theta#

#=(cos^2theta-sin^2theta)/(cos^2theta+sin^2theta)#

#=cos^2theta-sin^2theta=cos2theta#