Question

If a sin theta + b cos theta = c , then prove that a cos theta - b sin theta = the whole under root a 2 + b 2 - c 2 . plz answer soon if u want a thumbzzzz up guys !!!!!!!!!!!!!

Answer 1

Assuming theta = A for clarity

We are given,

=> a sinA + b cosA = c

Squaring both sides :-

=> (a sinA + b cosA)² = c²
=> a² sin²A + b² cos²A + 2ab sinA.cosA = c²
=> a²(1-cos²A) + b²(1-sin²A) + 2ab sinA.cosA = c²
=> a²-a²cos²A +b²-b²sin²A + 2ab sinA.cosA = c²
=>-a²cos²A -b²sin²A +2ab sinA.cosA = c²- a²- b²
=> a²cos²A +b²sin²A -2ab sinA.cosA = a²+ b²-c²
=> (a cosA - b sinA)² = a² + b²-c²
Taking square root on both sides :-
=> a cosA - b sinA =
$\sqrt[ + - ]{a {}^{2} + b {}^{2} - c {}^{2} }$
Q.E.D.