Question

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

Answer 1

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`

Answer 2

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)`

Answer 3

`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`