Question

25) if 3 Sin 0 + 4COS 0 =5
then the value of 4sin 0
-3cos 0​

Answer 1

Answer:

ANSWER

We have,

3sinθ+4cosθ=5 ………. (1)

On squaring both sides, we get

(3sinθ+4cosθ)

2

=5

2

9sin

2

θ+16cos

2

θ+24sinθcosθ=25

9(1−cos

2

θ)+16(1−sin

2

θ)+12×2sinθcosθ=25

9−9cos

2

θ+16−16sin

2

θ+12×2sinθcosθ=25

25−9cos

2

θ−16sin

2

θ+12×2sinθcosθ=25

9cos

2

θ+16sin

2

θ−12×2sinθcosθ=0

(3cosθ−4sinθ)

2

=0

3cosθ−4sinθ=0

Hence, the value is 0.