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Question
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
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Solution
LHS = `(1 + cosA)/(1 - cosA)`
= `(1 + 1/secA)/(1 - 1/secA) = (secA + 1)/(secA - 1)`
= `(secA + 1)/(secA - 1) . (secA - 1)/(secA - 1)`
= `(sec^2A - 1)/(secA - 1)^2 = tan^2A/(secA - 1)^2` (`Q sec^2A - 1 = tan^2A`)
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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`
