Sec thita -1÷ sec thita +1 =(sin thita ÷ 1+ costhita)^2 ?

3 Answers
May 22, 2018

Please see the proof below

Explanation:

We need

sectheta=1/costheta

sin^2theta+cos^2theta=1

Therefore the

LHS=(sectheta-1)/(sectheta+1)

=(1/costheta-1)/(1/costheta+1)

=(1-costheta)/(1+costheta)

=((1-costheta)(1+costheta))/((1+costheta)(1+costheta))

=(1-cos^2theta)/(1+costheta)^2

sin^2theta/(1+costheta)^2

=(sintheta/(1+costheta))^2

=RHS

QED

May 22, 2018

LHS=(secx-1)/(secx+1)

=(1/cosx-1)/(1/cosx+1)

=(1-cosx)/(1+cosx)*[(1+cosx)/(1+cosx)]

=(1-cos^2x)/(1+cosx)^2=sin^2x/(1+cosx)^2=(sinx/(1+cosx))^2=RHS

May 22, 2018

Explanation in below

Explanation:

(secx-1)/(secx+1)

=((secx-1)*(secx+1))/(secx+1)^2

=((secx)^2-1)/(secx+1)^2

=(tanx)^2/(secx+1)^2

=(sinx/cosx)^2/(1/cosx+1)^2

=((sinx)^2/(cosx)^2)/((1+cosx)^2/(cosx)^2)

=(sinx)^2//(1+cosx)^2

=(sinx/(1+cosx))^2