Question

If a cosΘ - b sinΘ = c, then a sinΘ + b cosΘ is equal to
(a) ±√a²+b²+c²
(b) ±√a²+b²-c²
(c) ±√c²-a²-b²
(d) 1/3

Answer 1

Step-by-step explanation:

Question:- a cosΘ - b sinΘ = c, then a sinΘ + b cosΘ is equal to

_________

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 ) .

hope this helps you ☺️

Answer 2

Answer:

Correct option is

A

±

a

2

+b

2

+c

2

Given, a cos θ - b sin θ = c

Squaring both sides,

a² cos² θ + b² sin² θ - 2ab sin θ cos θ = c²

∴ a² ( 1 - sin² θ ) + b² ( 1 - cos² θ ) - 2ab sin θ cos θ = c²

∴ a² - a² sin² θ + b² - b² cos² θ - 2ab sin θ cos θ = c²

∴ a² + b² - c² = a² sin² θ + b² cos² θ + 2ab sin θ cos θ

∴ ( √(a²+b²-c²) )² = ( a sin θ + b cos θ )²

∴ a sin θ + b cos θ = ± √(a²+b²-c²)