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Question
Prove that:
(secθ - cosθ)(cotθ + tanθ) = tanθ.secθ.
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Solution 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.
Solution 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.
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Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)
= `(1/square - cos θ) (square/square + square/square)` ......`[∵ sec θ = 1/square, cot θ = square/square and tan θ = square/square]`
= `((1 - square)/square) ((square + square)/(square square))`
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= `square/(square square)`
= tan θ.sec θ
= R.H.S.
∴ L.H.S. = R.H.S.
∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ
