How do you simplify #(1 + sec theta)/sec theta = sin^2 theta/(1 - cos theta)#?

1 Answer
Nov 27, 2015

Please look at the explanation

Explanation:

Remember: The trigonometric identities '

#sin^2theta+cos^2theta= 1# #hArr# #sin^2theta= 1-cos^2theta#

#1/costheta= sectheta# #hArr# #1/sectheta= costheta#

#(1+sectheta)/sectheta = (sin^2theta)/(1-costheta)#

#=># If we start from the right hand side, multiply by the conjugate of the denominator

#(1+sectheta)/sectheta = [(sin^2theta)/(1-costheta)]color(red)([(1+costheta)/(1+costheta)]#

#(1+sectheta)/sectheta = ((sin^2theta)(1+costheta)]/(1-cos^2theta)#

#(1+sectheta)/sectheta = [(sin^2theta)(1+costheta)]/(sin^2theta)#

#(1+sectheta)/sectheta = [cancel(sin^2theta) (1+costheta)]/(cancel(sin^2theta)#

#(1+sectheta)/sectheta = 1+costheta#

#(1+sectheta)/sectheta = 1+1/(sectheta)# common denominator

#(1+sectheta)/sectheta = 1*[color(red)sectheta]/color(red)sectheta+1/(sectheta)#

#(1+sectheta)/sectheta = (1+sectheta)/sectheta#

Done!