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

Jan 17, 2018

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))

$= \frac{1 - \cos \beta}{\sin} \beta$

$= \frac{1}{\sin} \beta - \cos \frac{\beta}{\sin} \beta$

$= \csc \beta - \cot \beta$

2nd part of LHS

=(1+sinβ+cosβ)/(1+sinβ-cosβ)

=Reciprocal of 1st part of LHS

$= \frac{1}{\csc \beta + \cot \beta}$

$= \frac{\csc \theta - \cot \theta}{{\csc}^{2} \beta - {\cot}^{2} \beta}$

$= \csc \beta - \cot \beta$

LHS
 = cscβ-cotbeta+cscbeta+cotbeta

$= 2 \csc \beta = R H S$

Jan 18, 2018

Proving this identity in explanation section.

#### Explanation:

After using $u = \beta$, this equation became

$\frac{1 + \sin u - \cos u}{1 + \sin u + \cos u} + \frac{1 + \sin u + \cos u}{1 + \sin u - \cos u}$

=$\frac{{\left(1 + \sin u - \cos u\right)}^{2} + {\left(1 + \sin u + \cos u\right)}^{2}}{\left(1 + \sin u + \cos u\right) \cdot \left(1 + \sin u - \cos u\right)}$

After expanding equation, numerator became

$A = 1 + {\left(\sin u\right)}^{2} + {\left(\cos u\right)}^{2} + 2 \sin u - 2 \cos u - 2 \sin u \cdot \cos u + 1 + {\left(\sin u\right)}^{2} + {\left(\cos u\right)}^{2} + 2 \sin u + 2 \cos u + 2 \sin u \cdot \cos u$

=$2 + 2 \cdot \left[{\left(\sin u\right)}^{2} + {\left(\cos u\right)}^{2}\right] + 4 \sin u$

=$2 + 2 + 4 \sin u$

=$4 + 4 \sin u$

=$4 \cdot \left(1 + \sin u\right)$

Also denominator became,

$B = {\left(1 + \sin u\right)}^{2} - {\left(\cos u\right)}^{2}$

=$1 + {\left(\sin u\right)}^{2} + 2 \sin u - {\left(\cos u\right)}^{2}$

=${\left(\sin u\right)}^{2} + {\left(\cos u\right)}^{2} + {\left(\sin u\right)}^{2} + 2 \sin u - {\left(\cos u\right)}^{2}$

=$2 \sin u + 2 {\left(\sin u\right)}^{2}$

=$2 \sin u \cdot \left(1 + \sin u\right)$

Thus,

$\frac{1 + \sin u - \cos u}{1 + \sin u + \cos u} + \frac{1 + \sin u + \cos u}{1 + \sin u - \cos u}$

=$\frac{A}{B}$

=$\frac{4 \cdot \left(1 + \sin u\right)}{2 \sin u \cdot \left(1 + \sin u\right)}$

=$\frac{2}{\sin} u$

=$2 \csc u$

=$2 \csc \left(\beta\right)$

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)))

$= \frac{2}{\sin} \beta = 2 \csc \beta$