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प्रश्न
If sinA + cosA = `sqrt(2)` , prove that sinAcosA = `1/2`
उत्तर
We Know , `(sinA + cosA)^2 = sin^2A + cos^2A + 2sinA.cosA`
Given , (sinA + cosA) = `sqrt(2)`
⇒ 2 = 1 + 2sinA.cosA
⇒ 2sinA.cosA = 1
⇒ sinA.cosA = `1/2`
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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`
