Question

If a sin theta + b cos theta = c , then prove that a cos theta - b sin theta = the whole under root a 2 + b 2 - c 2 . plz answer soon if u want a thumbzzzz up guys !!!!!!!!!!!!!

Answer 1

asinθ + bcosθ = c
taking square both sides,
(asinθ + bcosθ)² = c²
⇒a²sin²θ + b²cos²θ + 2absinθ.cosθ = c² --------(1)

Let acosθ - bsinθ = x
Squaring both sides
(acosθ - bsinθ)² = x²
⇒a²cos²θ + b²sin²θ -2absinθ.cosθ = x² ------(2)

Add equation (1) and (2),
a²sin²θ + b²cos²θ +2abinθ.cosθ + a²cos²θ + b²sin²θ -2absinθ.cosθ = c² + x²
⇒(a² + b²)cos²θ + (a² +b²)sin²θ = c² + x²
⇒(a² + b²)[sin²θ + cos²θ ] = c² + x²
⇒(a² + b²) = c² + x² [∵ sin²x + cos²x = 1 ]
⇒(a² + b² - c²) = x²
Take square root both sides,
$\bold{\pm\sqrt{a^2 + b^2 - c^2} =x}$
Hence, acosθ - bsinθ = $\bold{\sqrt{a^2 + b^2 - c^2}}$

Answer 2

Given :-

→ a sin∅ + b cos∅ = c .......(1) .

Now,

→ ( a sin∅ + b cos∅ )² + ( a cos∅ - b sin∅ )² .

= a²sin²∅ + b²cos²∅ + 2a sin∅ b cos∅ + a²cos²∅ + b²sin²∅ - 2a sin∅ b cos∅ .

= a²sin²∅ + a²cos²∅ + b²cos²∅ + b²sin²∅ .

= a²( sin²∅ + cos²∅ ) + b²( cos²∅ + sin²∅ ) .

= a² + b² . [ ∵ sin²∅ + cos²∅ = 1 ] .

Thus, ( a sin∅ + b cos∅ )² + ( a cos∅ - b sin∅ )² = ( a² + b² ) .

⇒ c² + ( a cos∅ - b sin∅ )² = ( a² + b² ) .

⇒ ( a cos∅ - b sin∅ )² = ( a² + b² - c² ) .

⇒ ( a cos∅ - b sin∅ ) = √( a² + b² - c² ) .

Hence, $\sf \pink { (a \: { \cos }^{2} \theta - b \: { \sin }^{2} \theta ) = \sqrt{ {a}^{2} + {b}^{2} - {c}^{2} } }.$