Question

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

Answer 1

$\underline{\underline{\Huge{\textbf{Question:}}}}$

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



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

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

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



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

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



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

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

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



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

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

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

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



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

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

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



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

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



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

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

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



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

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

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

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



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

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

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

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

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