Question

if 3cos alpha - 4sin alpha =5 then prove that 3cos alpha + 4cos alpha = 0 ​

Answer 1

We have,

3sinθ+4cosθ=5 ………. (1)

On squaring both sides, we get

(3sinθ+4cosθ) ^2=5 ^2

9sin ^2θ+16cos ^2θ+24sinθcosθ=25

9(1−cos ^2θ)+16(1−sin^2θ)+12×2sinθcosθ=25

9−9cos ^2θ+16−16sin^2θ+12×2sinθcosθ=25

25−9cos^2θ−16sin^2θ+12×2sinθcosθ=25

9cos^2θ+16sin^2θ−12×2sinθcosθ=0

(3cosθ−4sinθ)^2=0

3cosθ−4sinθ=0

Hence, the value is 0.

Answer 2

Answer:

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We have,

3sinθ+4cosθ=5 ………. (1)

On squaring both sides, we get

(3sinθ+4cosθ) ^2=5 ^2

9sin ^2θ+16cos ^2θ+24sinθcosθ=25

9(1−cos ^2θ)+16(1−sin^2θ)+12×2sinθcosθ=25

9−9cos ^2θ+16−16sin^2θ+12×2sinθcosθ=25

25−9cos^2θ−16sin^2θ+12×2sinθcosθ=25

9cos^2θ+16sin^2θ−12×2sinθcosθ=0

(3cosθ−4sinθ)^2=0

3cosθ−4sinθ=0

Hence, the value is 0.

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