If3sin–4cos=0.Findthevaluesoftan,sec.
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If 3sinθ+4cosθ=5, then find the value of 4sinθ−3cosθ.
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.
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