Question

pls answer it quickly ​

Answer 1

Answer:

Let d = a sin θ - b cos θ.

A useful identity is:

( u² + v² ) ( x² + y² ) = ( ux + vy )² + ( uy - vx )²

To check it is true, expand both sides; they both become

u²x² + u²y² + v²x² + v²y²

Putting in u = a, v = b, x = cos θ, y = sin θ, this gives:

( a² + b² ) ( cos²θ + sin²θ ) = c² + d²

=> d² = a² + b² - c²

=> d = ±√( a² + b² - c² )

Answer 2

Answer:

Step-by-step explanation:

if a cosФ-b sinФ=+√a^2+b^2-c^2

let a sinФ-b cosФ=x

squaring and adding in equation

a^2 cos^2Ф+b^2 sin'^2Ф+2 ab sinФcosФ=c^2

a^2 sin^2Ф+b^2 cos^2Ф-2 ab sinФcosФ=x^2

a^2(cos^2Ф+sin^2Ф)+b^2(sin^2Ф+cos^2Ф)=c^2+x^2

a^2+b^2-c^2=x^2

x=±√a^2+b^2-c^2

hence proved