A cos-b sin-m and a sin-b cos-n- then a2-b2

The correct option is D m^2+n^2Given: a cosθ+b sinθ=m and a sinθ b cosθ=n ⇒ m^2+n^2=(a cosθ+b sinθ)^2+(a sinθ b cosθ)^2 ⇒ m^2+n^2=a^2cos^2θ+b^2sin^2θ+2ab cosθ ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

a cosθ+b sinθ...

Question

$a cos θ + b sin θ = m a n d a sin θ - b cos θ = n, t h e n a^{2} + b^{2}$

m^{2} - n^{2}

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m^{2} n^{2}

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n^{2} - m^{2}

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m^{2} + n^{2}

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Solution

The correct option is D $m^{2} + n^{2}$
$G i v e n : a cos θ + b sin θ = m a n d a sin θ - b cos θ = n \Rightarrow m^{2} + n^{2} = (a cos θ + b sin θ)^{2} + (a sin θ - b cos θ)^{2} \Rightarrow m^{2} + n^{2} = a^{2} {cos}^{2} θ + b^{2} {sin}^{2} θ + 2 a b cos θ sin θ + a^{2} {sin}^{2} θ + b^{2} {cos}^{2} θ - 2 a b sin θ cos θ \Rightarrow m^{2} + n^{2} = a^{2} ({cos}^{2} θ + {sin}^{2} θ) + b^{2} ({sin}^{2} θ + {cos}^{2} θ) \Rightarrow m^{2} + n^{2} = a^{2} + b^{2} (∵ {sin}^{2} θ + {cos}^{2} θ = 1)$

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Similar questions

Q. If

a cos θ + b sin θ = m

and

a sin θ - b cos θ = n

, then

a^{2} + b^{2}

equals

Q. If a cos θ + b sin θ = m and a sin θ − b cos θ = n, prove that (m² + n²) = (a² + b²).

Q. If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a² + b² =

(a) m² − n²
(b) m²n²
(c) n² − m²
(d) m² + n²

Q. If

x = a cos θ + b sin θ

and

y = a sin θ - b cos θ

, then

a^{2} + b^{2} = x^{2} + y^{2}

Q. If

a cos θ + b sin θ = c

, then prove that:

a sin θ - b cos θ = \pm \sqrt{a^{2} + b^{2} - c^{2}}

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acosθ+bsinθ=m and asinθ−bcosθ=n, then a2+b2

The correct option is D m2+n2Given: a cosθ+b sinθ=m and a sinθ−b cosθ=n⇒m2+n2=(a cosθ+b sinθ)2+(a sinθ−b cosθ)2⇒m2+n2=a2cos2θ+b2sin2θ+2ab cosθ sinθ+a2sin2θ+b2cos2θ−2ab sinθcosθ⇒m2+n2=a2(cos2θ+sin2θ)+b2(sin2θ+cos2θ)⇒m2+n2=a2+b2 (∵sin2θ+cos2θ=1)

$a cos θ + b sin θ = m a n d a sin θ - b cos θ = n, t h e n a^{2} + b^{2}$