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प्रश्न
Prove the following identity :
secA(1 - sinA)(secA + tanA) = 1
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उत्तर
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`
