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Question
Prove that:
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
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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.
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