If a SinA + b cosA = c, then a cosA - b sinA is equal to:1. \(\sqrt {{

1.	If a SinA + b cosA = c, then a cosA - b sinA is equal to:1. \(\sqrt {{a^2} + {b^2} + {c^2}}\)2. \(\sqrt {{a^2} - {b^2} + {c^2}}\)3. \(\sqrt {{a^2} + {b^2} - {c^2}}\)4. \(\sqrt {{a^2} - {b^2} - {c^2}}\)
Answer» Correct Answer - Option 3 : \(\sqrt {{a^2} + {b^2} - {c^2}}\) GIven: a SinA + b cosA = c Formula used: If a Sinθ + b Cosθ = c Then, a Cosθ – b Sinθ = \(\sqrt {{a^2} + {b^2} - {c^2}}\) Calculation: Apply above formula on a SinA + b cosA = c We get ⇒ a CosA – b SinA = \(\sqrt {{a^2} + {b^2} - {c^2}}\) ∴ a CosA – b SinA is equla to \(\sqrt {{a^2} + {b^2} - {c^2}}\)

If a SinA + b cosA = c, then a cosA - b sinA is equal to:1. \(\sqrt {{a^2} + {b^2} + {c^2}}\)2. \(\sqrt {{a^2} - {b^2} + {c^2}}\)3. \(\sqrt {{a^2} + {b^2} - {c^2}}\)4. \(\sqrt {{a^2} - {b^2} - {c^2}}\)

Answer» Correct Answer - Option 3 : \(\sqrt {{a^2} + {b^2} - {c^2}}\)

GIven:

a SinA + b cosA = c

Formula used:

If a Sinθ + b Cosθ = c

Then, a Cosθ – b Sinθ = \(\sqrt {{a^2} + {b^2} - {c^2}}\)

Calculation:

Apply above formula on a SinA + b cosA = c

We get

⇒ a CosA – b SinA = \(\sqrt {{a^2} + {b^2} - {c^2}}\)

∴ a CosA – b SinA is equla to \(\sqrt {{a^2} + {b^2} - {c^2}}\)