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Question
Prove the following identity :
secA(1 + sinA)(secA - tanA) = 1
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Solution
LHS = secA(1 + sinA)(secA - tanA)
= `1/cosA(1 + sinA)(1/cosA - sinA/cosA)`
= `((1 + sinA))/cosA((1-sinA)/cosA) = (1-sin^2A)/cos^2A`
= `(cos^2A/cos^2A) = 1` = RHS
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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`
