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Question
Prove the following:
`(tantheta + sectheta - 1)/(tantheta + sectheta + 1) = tantheta/(sec theta + 1)`
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Solution 1
We know that,
tan2θ = sec2θ – 1
∴ tanθ.tanθ = (secθ + 1)(secθ – 1)
∴ `tantheta/(sec theta + 1) = (sectheta - 1)/tantheta`
By the theorem on equal ratios, we get
`tantheta/(sec theta + 1) = (sectheta - 1)/tantheta = (tantheta + sectheta - 1)/(tantheta + 1 + tantheta)`
∴ `(tantheta + sectheta - 1)/(tantheta + sectheta + 1) = tantheta/(sec theta + 1)`
Solution 2
`(tantheta + sectheta - 1)/(tantheta + sectheta + 1) = tantheta/(sec theta + 1)`
(tanθ + secθ − 1)(secθ + 1) = tanθ (tanθ + secθ + 1)
tanθ secθ + tanθ + sec2θ − 1
tan2θ + tanθ secθ + tanθ
tanθ secθ + tanθ + sec2θ − 1 = tan2θ + tanθ secθ + tanθ
`(tantheta + sectheta - 1)/(tantheta + sectheta + 1) = tantheta/(sec theta + 1)`
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