Question

$a \cosθ- b\sinθ = c \\ then \: prove \: that \: \\ a \sinθ + b \cosθ = \sqrt{ {a }^{2} + {b}^{2} + {c}^{2} } \\ if \: any \: body \: prove \: it \: i \: will \: mark \: them \: into \: brainlst$
but it's my challenge to all nobody can prove can prove this ques ​

Answer 1

GIVEN :–

$\\ \to \sf a \cos( \theta) - b \sin( \theta) = c$

TO PROVE :–

$\\ \to \sf a \sin\theta + b \cos\theta = \pm \sqrt{ {a }^{2} + {b}^{2} - {c}^{2} }$

SOLUTION :–

• Let's take a expression –

$\\ \implies \sf (a \cos \theta-b \sin \theta)^{2} + (a \sin \theta + b \cos \theta)^{2} = {(a \cos (\theta)) }^{2} + {(b \sin( \theta))}^{2} - 2ab \sin( \theta) \cos( \theta) + {(a \sin (\theta)) }^{2} + {(b \cos( \theta))}^{2} + 2ab \sin( \theta) \cos( \theta)$

$\\ \implies \sf (c)^{2} + (a \cos \theta + b \sin \theta)^{2} = {(a \cos (\theta)) }^{2} + {(b \sin( \theta))}^{2} + {(a \sin( \theta) )}^{2} + {(b \cos( \theta))}^{2}$

$\\ \implies \sf (c)^{2} + (a \sin \theta + b \cos \theta)^{2} = {a}^{2}( { \sin}^{2} \theta + { \cos}^{2} \theta ) + {b}^{2}( { \cos}^{2} \theta +{ \sin}^{2} \theta)$

$\\ \implies \sf (c)^{2} + (a \sin \theta + b \cos \theta)^{2} = {a}^{2} + {b}^{2}$

$\\ \implies \sf (a \sin \theta + b \cos \theta)^{2} = {a}^{2} + {b}^{2} - {c}^{2}$

• Take square root on both sides –

$\\ \implies \sf a \sin \theta + b \cos \theta = \pm \sqrt{ {a}^{2} + {b}^{2} - {c}^{2}}$

$\\ \:\: \longrightarrow \:\: \large { \boxed {{ \sf Hence \: \: proved }}}$

USED FORMULA :–

$\\ \sf \: (1) \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

$\\ \sf \: (2) \: {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab$

$\\ \sf \: (3) \: { \sin }^{2} \theta + { \cos}^{2} \theta = 1$