Prove that
2cos3θ−cosθsinθ−2sin3θ=tanθ
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Answer:
Answer
Given,
2cos
3
θ−cosθ
sinθ−2sin
3
θ
=tanθ
L.H.S=
2cos
3
θ−cosθ
sinθ−2sin
3
θ
=
cosθ(2cos
2
θ−1)
sinθ(1−2sin
2
θ)
=
cosθ
sinθ
[
2(1−sin
2
θ)−1
1−2sin
2
θ
]
=tanθ[
2−2sin
2
θ−1
1−2sin
2
θ
]
=tanθ[
1−2sin
2
θ
1−2sin
2
θ
]
=tanθ
=R.H.S
∴
2cos
3
θ−cosθ
sinθ−2sin
3
θ
=tanθ
Step-by-step explanation:
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