Question

Prove that

`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`

Answer 1

`"LHS" = sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ))`

Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get,

`"LHS" = sqrt((1 + sin θ)/(1 - sin θ) × (1 + sin θ)/(1 + sin θ)) + sqrt((1 - sin θ)/(1 + sin θ) × (1 - sin θ)/(1 - sin θ))`

`"LHS" = sqrt((1 + sin θ)^2/(1 - sin^2 θ)) + sqrt((1 - sin θ)^2/(1 - sin^2 θ))`

`"LHS" = sqrt((1 + sin^2θ)/(1 - sin^2 θ)) + sqrt((1 - sin^2θ)/(1 - sin^2 θ))`

`"LHS" = sqrt((1 + sin^2θ)/(cos^2 θ)) + sqrt((1 - sin^2θ)/(cos^2 θ))`

`"LHS" = (1 + sin θ)/(cos θ) + (1 - sin θ)/(cos θ)`

`"LHS" = (1 + cancel(sin θ) + 1 -cancel(sin θ))/(cos θ)`

LHS = `2/(cos θ)`

LHS = 2. `1/(cos θ)`

LHS = 2. sec θ

RHS = 2. sec θ

LHS = RHS

Hence proved.