Advertisements
Advertisements
Question
Prove that:
(secθ - cosθ)(cotθ + tanθ) = tanθ.secθ.
Advertisements
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.
APPEARS IN
RELATED QUESTIONS
If \[\sin\theta = \frac{7}{25}\], find the values of cosθ and tanθ.
If 5 secθ – 12 cosecθ = 0, find the values of secθ, cosθ, and sinθ.
If tanθ = 1 then, find the value of
`(sinθ + cosθ)/(secθ + cosecθ)`
Prove that:
cos2θ (1 + tan2θ)
Prove that:
Prove that:
Prove that:
Choose the correct alternative answer for the following question.
cosec 45° =?
Choose the correct alternative answer for the following question.
1 + tan2 \[\theta\] = ?
Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ
Prove the following.
Prove the following.
\[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]
Prove the following.
If sinθ = `8/17`, where θ is an acute angle, find the value of cos θ by using identities.
Show that:
`sqrt((1-cos"A")/(1+cos"A"))=cos"ecA - cotA"`
In ΔPQR, ∠P = 30°, ∠Q = 60°, ∠R = 90° and PQ = 12 cm, then find PR and QR.
Prove that: (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ
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))`
= `square/square xx 1/(square square)` ......`[(∵ square + square = 1),(∴ square = 1 - square)]`
= `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
