If a sinθ + b cosθ = c then the value of a cos θ - b sin θ is :

a sinθ + b cosθ = c

on squaring both sides , we get

$$a^2 sin^2 \theta$$ + $$b^2 cos^2 \theta$$ + 2 a b $$sin \theta cos \theta$$ = $$c^2$$ ...............(1)

We need to find value of : a sinθ - b cosθ.....(2)

So, on squaring both sides of equation 2,we get

( a cosθ - b sinθ)$$^2$$ = $$a^2 cos^2 \theta$$ + $$b^2 sin^2 \theta$$ - 2 a b $$cos \theta sin \theta$$......(3)

Adding equation 1 and 3, we can say

c$$^2$$ + (a cos θ - b sin θ)$$^2$$ = $$a^2 cos^2 \theta$$ + $$b^2 sin^2 \theta$$ + $$a^2 sin^2 \theta$$ + $$b^2 cos^2 \theta$$

c$$^2$$ + (a cos θ - b sin θ)$$^2$$ = $$a^2 cos^2 \theta$$ + $$b^2 sin^2 \theta$$ + $$a^2 sin^2 \theta$$ + $$b^2 cos^2 \theta$$

c$$^2$$ + (a cos θ - b sin θ)$$^2$$ = $$a^2 cos^2 \theta$$ + $$a^2 sin^2$$ + $$b^2 cos^2 \theta$$ + $$b^2 sin^2$$

c$$^2$$ + ( a cos θ - b sin θ)$$^2$$ = $$a^2 + b^2 $$

(a cos θ - b sin θ)$$^2$$ = $$a^2 + b^2 $$ - c$$^2$$

( a cos θ - b sin θ) = $$\pm\sqrt{a^{2}+b^{2}-c^{2}}$$