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प्रश्न
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that a2 + b2 = m2 + n2
उत्तर
R.H.S `m^2 sin^2 theta`
`= (a cos theta + b sin theta)^2 + (a sin theta - b cos theta)^2`
`= a^2 cos^2 theta + b^2 sin^2 theta + 2 ab sin theta cos theta + a^2 sin^2 theta + b^2 cos^2 theta - 2 ab sin theta cos theta`
`= a^2 cos^2 theta + b^2 cos^2 theta + b^2 sin^2 theta + a^2 sin^2 theta`
`= a^2(sin^2 theta + cos^2 theta) + b^2(sin^2 theta + cos^2 theta)`
`=a^2 + b^2` (∵ `sin^2 theta + cos^2 theta = 1`)
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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`
