Question

if acos theta - bsin theta =c then prove that asin theta +bcostheta=+-√a²+√b²-√c²

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$\huge\underline{\colorbox{pink}{Answer}}$

Given :-

→ a cos∅ + b sin∅ = c ...(1) .

Now,

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

= a²cos²∅ + b²sin²∅ - 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 cos∅ - b sin∅ )² + ( a sin∅ + b cos∅ )² = ( a² + b² ) .

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

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

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

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