How do you prove the identity #[1-sinx]/secx = (cos^3x)/(1+sinx)#?

Question

#[1-sinx]/secx = (cos^3x)/(1+sinx)#

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Answer 1 · 2018-04-07T10:42:31

After editing the question:
#(1-sinx)/secx =color(red) (cos3x)/(1+sinx)to (1-sinx)/secx = color(red)(cos^3x)/(1+sinx)#

We know that,

#color(green)((1)sin^2theta+cos^2theta=1=>cos^2theta=1- sin^2theta#

#color(green)((2)sectheta=1/costheta#

We have to prove,

#(1-sinx)/secx = (cos^3x)/(1+sinx)#

We take

#LHS=(1-sinx)/secx #

#=(1-sinx)/secx xx(1+sinx)/(1+sinx)#

#=1/(secx)xx(1-sin^2x)/(1+sinx)...toApply (1) and (2)#

#=cosx*cos^2x/(1+sinx)#

#=cos^3x/(1+sinx)#

#=RHS#