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Question
Prove that:
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
Solution 1
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
RHS = `(tanθ + secθ + 1)/(tanθ + secθ - 1)`
`"RHS" = (tanθ + secθ + 1)/((tanθ + secθ) - 1) × (tanθ + secθ + 1)/((tanθ + secθ) + 1) ...("On rationalising the denominator")`
`"RHS" = ((tanθ + secθ + 1)^2)/((tanθ + secθ)^2 - 1)`
`"RHS" = (tan^2θ + sec^2θ + 1 + 2tanθsecθ + 2tanθ + 2secθ)/(tan^2θ + 2tanθ.secθ + sec^2θ - 1) ...{((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc),((a + b)^2 = a^2 + 2ab + b^2):}`
`"RHS" = ((1 + tan^2θ) + sec^2θ + 2tanθsecθ + 2tanθ + 2secθ)/(tan^2θ + 2tanθ.secθ + (sec^2θ - 1))`
`"RHS" = (sec^2θ + sec^2θ + 2tanθsecθ + 2tanθ + 2secθ)/(tan^2θ + 2tanθ.secθ + tan^2θ)`
`"RHS" = (2sec^2θ + 2tanθsecθ + 2tanθ + 2secθ)/(2tan^2θ + 2tanθ.secθ)`
`"RHS" = [2secθ (tanθ + secθ) + 2(tanθ + secθ)]/[2tanθ(tanθ + secθ)]`
`"RHS" = [(2secθ + 2)(cancel(tanθ + secθ))]/[2tanθ(cancel(tanθ + secθ))]`
`"RHS" = [cancel2(secθ + 1)]/[cancel2(tanθ)]`
`"RHS" = (secθ + 1)/(tanθ)`
`"RHS" = (secθ + 1)/(tanθ) × (secθ - 1)/(secθ - 1)`
`"RHS" = (sec^2θ - 1)/(tanθ(secθ - 1)) ...[(a - b)(a + b) = a^2 - b^2]`
`"RHS" = (tan^cancel2θ)/(canceltanθ(secθ - 1))`
`"RHS" = tanθ/(secθ - 1)`
LHS = `"tanθ"/("secθ" – 1)`
LHS = RHS
Hence proved.
Solution 2
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
LHS = `"tanθ"/("secθ" – 1)`
`"LHS" = "tanθ"/("secθ" – 1) × ("secθ" + 1)/("secθ"+ 1) ...("On rationalising the denominator")`
`"LHS" = ("tanθ"("secθ" + 1))/("sec"^2θ" - 1) ...[(a + b)(a - b) = a^2 - b^2]`
`"LHS" = ("tanθ"("secθ" + 1))/("tan"^2θ") ...{(∵ 1 + tan^2θ = sec^2θ),(∵ sec^2θ - 1 = tan^2θ):}`
`"LHS" = ("secθ" + 1)/"tanθ"`
∴ `"tanθ"/("secθ" – 1) = ("secθ" + 1)/"tanθ"`
∴ By theorem on equal ratios,
`"tanθ"/("secθ" – 1) = ("secθ" + 1)/"tanθ" = (tanθ + (secθ + 1))/((tanθ) + secθ - 1)`
`"LHS" = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
`"RHS" = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
LHS = RHS
Hence proved.
RELATED QUESTIONS
Prove the following trigonometric identities.
`tan^2 theta - sin^2 theta tan^2 theta sin^2 theta`
Prove the following trigonometric identity.
`(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`
If cos θ + cos2 θ = 1, prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1
Prove the following identities:
`1/(1 + cosA) + 1/(1 - cosA) = 2cosec^2A`
If sin A + cos A = m and sec A + cosec A = n, show that : n (m2 – 1) = 2 m
Prove the following identities:
`cosA/(1 + sinA) + tanA = secA`
Prove the following identities:
`cot^2A((secA - 1)/(1 + sinA)) + sec^2A((sinA - 1)/(1 + secA)) = 0`
Prove that:
`cosA/(1 + sinA) = secA - tanA`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
`(sec theta + tan theta )/( sec theta - tan theta ) = ( sec theta + tan theta )^2 = 1+2 tan^2 theta + 25 sec theta tan theta `
`(1+ cos theta + sin theta)/( 1+ cos theta - sin theta )= (1+ sin theta )/(cos theta)`
Write the value of `4 tan^2 theta - 4/ cos^2 theta`
If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
Prove the following identity :
`sqrt((1 + sinq)/(1 - sinq)) + sqrt((1- sinq)/(1 + sinq))` = 2secq
Prove the following identity :
`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2 = 2((1 + sin^2θ)/(1 - sin^2θ))`
For ΔABC , prove that :
`tan ((B + C)/2) = cot "A/2`
Without using a trigonometric table, prove that
`(cos 70°)/(sin 20°) + (cos 59°)/(sin 31°) - 8sin^2 30° = 0`.
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= R.H.S
If sinA + sin2A = 1, then the value of the expression (cos2A + cos4A) is ______.
sec θ when expressed in term of cot θ, is equal to ______.
