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Question
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
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Solution
Given: sin θ + 2 cos θ = 1
Squaring on both sides,
(sin θ + 2 cos θ)2 = 1
⇒ sin2 θ + 4 cos2 θ + 4sin θ cos θ = 1
Since, sin2 θ = 1 – cos2 θ and cos2 θ = 1 – sin2 θ
⇒ (1 – cos2 θ) + 4(1 – sin2 θ) + 4sin θ cos θ = 1
⇒ 1 – cos2 θ + 4 – 4 sin2 θ + 4sin θ cos θ = 1
⇒ – 4 sin2 θ – cos2 θ + 4sin θ cos θ = – 4
⇒ 4 sin2 θ + cos2 θ – 4sin θ cos θ = 4
We know that,
a2 + b2 – 2ab = (a – b)2
So, we get,
(2sin θ – cos θ)2 = 4
⇒ 2sin θ – cos θ = 2
Hence proved.
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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`
