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Question
Prove the following identities:
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
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Solution
R.H.S. = `tan^2A/(secA - 1)^2`
= `(sec^2A - 1)/(secA - 1)^2` ...[sec2θ – tan2θ = 1 sec2θ – 1 = tan2θ]
= `((secA + 1)(secA - 1))/(secA - 1)^2`
= `(secA + 1)/(secA - 1)`
= `(1/(cosA) + 1)/(1/cosA - 1)`
= `((1 + cosA)/cosA)/((1 - cosA)/(cosA))`
= `(1 + cosA)/(1 - cosA)`
R.H.S. = L.H.S.
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
