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Question
If sin θ + sin2 θ = 1 show that: cos2 θ + cos4 θ = 1
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Solution
sin θ + sin2 θ = 1 ......[Given]
∴ sin θ = 1 − sin2 θ
∴ sin θ = cos2 θ ......[∵ 1 − sin2 θ = cos2 θ]
∴ sin2 θ = cos4 θ ......[Squaring both the sides]
∴ 1 − cos2 θ = cos4 θ ......[∵ sin2 θ = 1 − cos2 θ]
∴ 1 = cos2 θ + cos4 θ
∴ cos2 θ + cos4 θ = 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`
