Question

If 3 sin 0 + cos 0 = 2 then 3 cos 0 - sin 0 is =

(A) –√6

(B) √6

(C) √5

(D) -√5​

Answer 1

Answer:

We have,

3sinθ+4cosθ=5              ………. (1)

On squaring both sides, we get

(3sinθ+4cosθ)2=52

9sin2θ+16cos2θ+24sinθcosθ=25

9(1−cos2θ)+16(1−sin2θ)+12×2sinθcosθ=25

9−9cos2θ+16−16sin2θ+12×2sinθcosθ=25

25−9cos2θ−16sin2θ+12×2sinθcosθ=25

9cos2θ+16sin2θ−12×2sinθcosθ=0

(3cosθ−4sinθ)2=0

3cosθ−4sinθ=0

Hence, the value is 0.