Question

if a cosθ+b sinθ=c then (a sinθ-b cosθ)2 =?


A.c2-a2-b2

B.c2-a2+b2

C.a2-b2+c2

D.a2+b2-c2

Answer 1

given a cos theta + b sin theta = c
squaring on both sides we get
a² cos² theta + b² ²sin² theta + 2 ab cos theta sin theta = c²
a² (1-sin² theta) + b² (1-cos² theta) + 2ab cos theta sin theta = c²
a² - a² sin² theta + b² - b² cos² theta + 2 ab cos theta sin theta = c²
a² + b² - a² sin² theta - b² cos² theta + 2 ab cos theta sin theta = c²
a² + b² - (a sin theta - b cos theta )² = c²
⇒ (a sin theta - b cos theta )² = a ² + b² - c²