Question

Prove that:

(secθ - cosθ)(cotθ + tanθ) = tanθ.secθ.

Answer 1

(secθ - cosθ)(cotθ + tanθ) = tanθ  secθ.

LHS = (secθ - cosθ)(cotθ + tanθ)

`"LHS" =(1/cosθ - cosθ)(cosθ/sinθ + sinθ/cosθ)   ...{(secθ = 1/cosθ),(cot θ = cosθ/sinθ),(tan θ = sinθ/cosθ):}`

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

`"LHS" =((sin^2θ)/cosθ) × ((1)/(sinθcosθ))   ...{(cos^2θ + sin^2θ = 1),(∵ 1 - cos^2θ = sin^2θ):}`

`"LHS" =((sin^cancel2θ)/cosθ) × ((1)/(cancel(sinθ)cosθ))`

`"LHS" = sinθ/cosθ × 1/cosθ`

`"LHS" = tanθ.secθ   ...{(sinθ/cosθ = tanθ),(1/cosθ = secθ):}`

RHS = tanθ.secθ

LHS = RHS

Hence proved.

Answer 2

(secθ - cosθ)(cotθ + tanθ) = tanθ.secθ.

LHS = (secθ - cosθ)(cotθ + tanθ)

LHS = secθ(cotθ + tanθ) - cosθ(cotθ + tanθ)

LHS = secθ.cotθ + secθ.tanθ - cosθ.cotθ + cosθ.tanθ

`"LHS" = (1/cosθ)(cosθ/sinθ) + secθ.tanθ - cosθ.(cosθ/sinθ) - cosθ.(sinθ/cosθ)  ...{(secθ = 1/cosθ),(cotθ = cosθ/sinθ),(tanθ = sinθ/cosθ):}`

`"LHS" = (1/cancel(cosθ))(cancel(cosθ)/sinθ) + secθ.tanθ - cosθ.(cosθ/sinθ) - cancel(cosθ).(sinθ/cancel(cosθ))`

`"LHS" = 1/sinθ + secθ.tanθ - cos^2θ/sinθ - sinθ`

`"LHS" = 1/sinθ - cos^2θ/sinθ - sinθ + secθ.tanθ`

`"LHS" = (1- cos^2θ - sin^2θ)/sinθ + secθ.tanθ`

`"LHS" = (1- (cos^2θ + sin^2θ))/sinθ + secθ.tanθ`

`"LHS" = (1- 1)/sinθ + secθ.tanθ ....{cos^2θ + sin^2θ = 1:}`

`"LHS" = (0)/sinθ + secθ.tanθ`

LHS = 0 + secθ.tanθ

LHS = secθ.tanθ

RHS = secθ.tanθ

LHS = RHS

Hence proved.