Question

Heya ..
If a sinθ + b cosθ = c, then prove that a cosθ - b sinθ = √a²+b²-c² .

Follow my 2nd I'd Plz ​

Answer 1

$\underline\bold\red{AnswEr}$

Given , (a sinθ + b cosθ)2 + (a cosθ - b sinθ)2

= a2 sin2 θ b2 cos2 θ + 2ab sinθ cosθ + a2 cos2θ + b2 sin2 θ - 2ab sinθ cosθ

a2 (sin2 θ + cos2 θ ) + b2 (sin2 θ + cos2 θ )

= a2 +b2 ( ∵ sin2 θ + cos2 θ = 1 )

∴ (a sinθ + b cosθ)2 + (a sinθ - b cosθ)2 = a2 +b2

⇒ c2 + (a sinθ - b cosθ)2 = a2 +b2 (∵ a sinθ + b cosθ = c)

⇒ (a cosθ - b sinθ)2 = a2 +b2 - c2

⇒ a cosθ - b sinθ = ± √a2 +b2 - c2

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

$\displaystyle \: a sinθ + b cosθ = c$

Squaring both sides we get

$\displaystyle \: {( \: a sinθ + b cosθ\:) }^{2} \: = {c}^{2}$

$\implies \: \displaystyle \: {(a sinθ)}^{2} + \: 2 \times a sinθ \times \: b cosθ\: + {( \: b cosθ\:) }^{2} \: = {c}^{2}$

$\implies \: \displaystyle \: {a}^{2} {sin}^{2} θ + \: 2 ab sinθ cosθ\: + {b}^{2} {cos}^{2} θ \: = {c}^{2}$

$\implies \: \displaystyle \: {a}^{2} (1 - {cos}^{2} θ )+ \: 2 ab sinθ cosθ\: + {b}^{2} (1 - {sin}^{2} θ )\: = {c}^{2}$

$\implies \: \displaystyle \: {a}^{2} {cos}^{2} θ - \: 2 ab sinθ cosθ\: + {b}^{2} {sin}^{2} θ \: = {a}^{2} + {b}^{2} - {c}^{2}$

$\implies \: \displaystyle \: {( \: a cosθ - b sinθ\:) }^{2} \: = {a}^{2} + {b}^{2} - {c}^{2}$

$\displaystyle \: \therefore \: a cosθ - b sinθ = √a²+b²-c² .$

Hence proved