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प्रश्न
If `1 - cos^2θ = 1/4`, then θ = ?
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उत्तर
`1 - cos^2θ = 1/4` ...[Given]
∴ `sin^2θ = 1/4` ...`[(∵ sin^2θ + cos^2θ = 1),(∴ 1 - cos^2θ = sin^2θ)]`
∴ `sin θ = 1/2` ...[Taking square root of both sides]
∴ θ = 30° ...`[∵ sin 30^circ = 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`
