Math, asked by abhijeetdash321, 1 year ago

What is the proof????????????????????????????????????

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Answered by Avengers00
17
\underline{\underline{\Huge{\textbf{Question:}}}}

\mathsf{If asin\, \theta + bcos\, \theta = c, then}\\\mathsf{Prove \: that \: acos\, \theta - bsin\, \theta = \sqrt{a^{2}+b^{2}-c^{2}}}

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\underline{\underline{\Huge{\textbf{Solution:}}}}

\underline{\Large{\textsf{Step-1:}}}
\sf\textsf{Consider Given Equation}

\mathsf{asin\, \theta + bcos\, \theta = c} ———[1]

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\underline{\Large{\textsf{Step-2:}}}
\sf\textsf{Squaring on Both sides of Equation [1]}

\mathsf{\left(asin\, \theta + bcos\, \theta = c\right)^{2} = c^{2}}

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\underline{\Large{\textsf{Step-3:}}}
\sf\textsf{Using the Identity}

\bigstar \: \: \boxed{\displaystyle\Large{\mathbf{(a \: + \: b)^{2}\: = \: a^{2}\: +\: b^{2}\: +\: 2ab}}}

\implies \mathsf{a^{2}sin^{2}\, \theta + b^{2}cos^{2}\, \theta +2absin\, \theta\: cos\, \theta= c^{2}}

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\underline{\Large{\textsf{Step-4:}}}
\mathsf{Express\: sin^{2}\, \theta\: in\: terms\: of\: cos^{2}\, \theta}\\\mathsf{cos^{2}\, \theta\: in\: terms\: of\: sin^{2}\, \theta}

\sf\textsf{Using the Identity}
\bigstar \: \: \boxed{\displaystyle\Large{\mathbf{sin^{2}\, \theta\: + cos^{2}\, \theta \: = \: 1}}}

\implies \bf sin^{2}\, \theta = 1 - cos^{2}\, \theta
\implies \bf cos^{2}\, \theta = 1 - sin^{2}\, \theta

\implies \mathsf{a^{2}\left(1-cos^{2}\, \theta\right) + b^{2}\left(1-sin^{2}\, \theta\right) +2absin\, \theta\: cos\, \theta= c^{2}}

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\underline{\Large{\textsf{Step-5:}}}
\sf\textsf{Use Distributive property of Multiplication}

\bigstar\: \: \boxed{\displaystyle\Large{\mathbf{a(b \pm c)\: = \: ab\: \pm\: ac}}}

\implies\mathsf{a^{2}-a^{2}cos^{2}\, \theta+ b^{2}-b^{2}sin^{2}\, \theta +2absin\, \theta\: cos\, \theta= c^{2}}

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\underline{\Large{\textsf{Step-6:}}}
\sf\textsf{Keep the terms containing}\\\sf\textsf{sin and cos to one side of equation}

\implies \mathsf{-a^{2}cos^{2}\, \theta-b^{2}sin^{2}\, \theta +2absin\, \theta\: cos\, \theta= c^{2} - a^{2} - b^{2}}

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\underline{\Large{\textsf{Step-7:}}}
\sf\textsf{Take (-1) common from both sides}

\implies \mathsf{-1\left(a^{2}cos^{2}\, \theta+b^{2}sin^{2}\, \theta -2absin\, \theta\: cos\, \theta\right)= -1\left(a^{2} + b^{2}-c^{2}\right)}

\implies \mathsf{a^{2}cos^{2}\, \theta+b^{2}sin^{2}\, \theta -2absin\, \theta\: cos\, \theta= a^{2} + b^{2}-c^{2}} ———[2]

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\underline{\Large{\textsf{Step-8:}}}
\sf\textsf{Rewrite [2] using the Identity}

\bigstar \: \: \boxed{\displaystyle\Large{\mathbf{(a \: - \: b)^{2}\: = \: a^{2}\: +\: b^{2}\: -\: 2ab}}}

\sf\textsf{Here,}
\mathsf{a = acos\, \theta ; b = bsin\, \theta}

\implies \mathsf{\left(acos\, \theta-bsin\, \theta\right)^{2}= a^{2} + b^{2}-c^{2}}

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\underline{\Large{\textsf{Step-9:}}}
\sf\textsf{Apply Square root on both sides}

\implies \mathsf{\sqrt{\left(acos\, \theta-bsin\, \theta\right)^{2}}= \sqrt{a^{2} + b^{2}-c^{2}}}

\bigstar\: \boxed{\Large{\mathbf{\sqrt{x^{2}}\: =\: x}}}

\implies \mathbf{acos\, \theta-bsin\, \theta= \sqrt{a^{2} + b^{2}-c^{2}}}

\underline{\Large{\textsf{Hence Proved}}}

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abhijeetdash321: Thanks a lot
Avengers00: My pleasure (:
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