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प्रश्न
If a cos θ – b sin θ = c, then prove that (a sin θ + b cos θ) = `± sqrt("a"^2 + "b"^2 -"c"^2)`
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उत्तर
Given a cos θ – b sin θ = c
Squaring on both sides
(a cos θ – b sin θ)2 = c2
a2 cos2 θ + b2 sin2 θ – 2 ab cos θ sin θ = c2
a2 (1 – sin2 θ) + b2 (1 – cos2 θ) – 2 ab cos θ sin θ = c2
a2 – a2 sin2 θ + b2 – b2 cos2 θ – 2 ab cos θ sin θ = c2
– a2 sin2 θ – B2 – cos2 θ – 2 ab cos θ sin θ = – a2 – b2 + c2
a2 sin2 θ + b2 cos2 θ + 2 ab cos θ sin θ = a2 + b2 – c2
(a sin θ + b cos θ)2 – a2 + b2 – c2
a sin θ + b cos θ = `± sqrt("a"^2 + "b"^2 -"c"^2)`
Hence it is proved.
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