if asinA+bcosA=c,then prove that acosA-bsinA=root of …

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if asinA+bcosA=c,then prove that acosA-bsinA=root of a2+b2-c2

Posted by Ramesh N K Shetty 5 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 5 years, 9 months ago

We have, $asin\theta+bcos\theta=c$

On squaring both sides, we get

$(asin\theta+bcos\theta)^2=c^2$

(a sin θ)² + (b cos θ)² + 2(a sin θ) (b cos θ) = c²

⇒ a² sin² θ + b² cos² θ + 2ab sin θ cos θ = c²

⇒ a²(1 – cos² θ) + b² (1 – sin² θ) + 2 ab sin θ cos θ = c^{2 $[\because sin^2\theta+cos^2\theta=1]$}

⇒ a² – a² cos² θ + b² – b² sin² θ + 2ab sin θ cos θ = c²

⇒ –a² cos² θ – b² sin² θ + 2ab sin θ cos θ = c²– a² – b²

Taking Negative common,

⇒ a² cos² θ + b² sin² θ – 2ab sin θ cos θ = a² + b² – c²

⇒ (a cos θ)² + (b sin θ)² – 2(a cos θ) (b sin θ) = a²+ b² – c²

⇒ $(acos\theta-bsin\theta)^2=a^2+b^2-c^2$

⇒ $acos\theta-bsin\theta$ = $\pm \sqrt { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } }$

Hence proved, $acos\theta-bsin\theta$ = $\sqrt{a^2+b^2-c^2}$

1Thank You

ANSWER

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CBSE > Class 10 > Mathematics