Question

Ello!! ❤️


Question : Refer to the attachment !!

#No spam​

Answer 1

Refer to the attachment...

Answer 2

Step-by-step explanation:

Given :-

→ a cos∅ + b sin∅ = c .......(1) .

Now,

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

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

→ ( 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∅ .

→ ( 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∅ ) .

→ ( 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 ] .

→ ( 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 ) .

→ ( 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 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 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,proved