Question

(i) If a cose + b sino = m and a sino - b cose = n then
prove that a² + b2 = m² + n?​

Answer 1

$\huge\boxed{\fcolorbox{red}{yellow}{Shailesh}}$

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,

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Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

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,