Question

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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

given: a.sinθ+b.cosθ=c

To Prove: a.cosθ–bsinθ=±√(a²+b²–c²)

a.sinθ+b.cosθ=c (given)

on sq.ⁿ both sides:

(a.sinθ+b.cosθ)²=(c)²

or a².sin²θ+b².cos²θ+2a.sinθ.b.cosθ=c²

[∵sin²θ+cos²θ=1]

a²(1–cos²θ)+b²(1–sin²θ)+2sinθ.cosθ.ab=c²

a²–a.cos²θ+b²–b.sin²θ+2sinθ.cosθ.ab=c²

(a.cos²θ+b.sin²θ–2b.sinθ.a.cosθ)=a²+b²–c²

(a.cosθ–b.sinθ)²=a²+b²–c²

a.cosθ–b.sinθ=√(a²+b²–c²)=RHS

Hence, proved