Prove that (1+sinβ-cosβ)/(1+sinβ+cosβ) +(1+sinβ+cosβ)/(1+sinβ-cosβ)=2cscbeta?

3 Answers

1st part of LHS

=(1+sinβ-cosβ)/(1+sinβ+cosβ)

=(sinbeta(1+sinβ-cosβ))/(sinbeta(1+sinβ+cosβ))

=(sinbeta+sin^2β-sinbetacosβ)/(sinbeta(1+sinβ+cosβ))

=(sinbeta+1-cos^2β-sinbetacosβ)/(sinbeta(1+sinβ+cosβ))

=(sinbeta(1-cosbeta)+(1-cosβ)(1+cosβ))/(sinbeta(1+sinβ+cosβ))

=((1-cosbeta)(sinbeta+1+cosβ))/(sinbeta(1+sinβ+cosβ))

=(1-cosbeta)/sinbeta

=1/sinbeta-cosbeta/sinbeta

=cscbeta-cotbeta

2nd part of LHS

=(1+sinβ+cosβ)/(1+sinβ-cosβ)

=Reciprocal of 1st part of LHS

=1/(cscbeta+cotbeta)

=(csctheta-cottheta)/(csc^2beta-cot^2beta)

=cscbeta-cotbeta

Adding we get

LHS
= cscβ-cotbeta+cscbeta+cotbeta

=2cscbeta=RHS

Jan 18, 2018

Proving this identity in explanation section.

Explanation:

After using u=beta, this equation became

(1+sinu-cosu)/(1+sinu+cosu)+(1+sinu+cosu)/(1+sinu-cosu)

=[(1+sinu-cosu)^2+(1+sinu+cosu)^2]/[(1+sinu+cosu)*(1+sinu-cosu)]

After expanding equation, numerator became

A=1+(sinu)^2+(cosu)^2+2sinu-2cosu-2sinu*cosu+1+(sinu)^2+(cosu)^2+2sinu+2cosu+2sinu*cosu

=2+2*[(sinu)^2+(cosu)^2]+4sinu

=2+2+4sinu

=4+4sinu

=4*(1+sinu)

Also denominator became,

B=(1+sinu)^2-(cosu)^2

=1+(sinu)^2+2sinu-(cosu)^2

=(sinu)^2+(cosu)^2+(sinu)^2+2sinu-(cosu)^2

=2sinu+2(sinu)^2

=2sinu*(1+sinu)

Thus,

(1+sinu-cosu)/(1+sinu+cosu)+(1+sinu+cosu)/(1+sinu-cosu)

=A/B

=[4*(1+sinu)]/[2sinu*(1+sinu)]

=2/sinu

=2cscu

=2csc(beta)

Jan 19, 2018

LHS =(1+sinβ-cosβ)/(1+sinβ+cosβ) +(1+sinβ+cosβ)/(1+sinβ-cosβ)

=((1+sinβ-cosβ)^2+(1+sinβ+cosβ)^2)/((1+sinβ)^2-cos^2β)

=(2(1+sin^2β+cos^2β)+2sinβ-2sinbetacosβ-2cosbeta+2sinbeta+2sinbetacosbeta+2cosbeta)/((1+sin^2β+2sinbeta-cos^2β))

=(4+4sinβ)/((2sin^2β+2sinbeta)

=(cancel4^2(cancel(1+sinβ)))/(cancel2(sinbeta(cancel(sinβ+1)))

=2/sinbeta=2cscbeta