•Prove that—
sin A - 2sin^3 A/2cos^3 A - cos A = tan A
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By solving LHS we get
⇒
cosA(2cos
2
A−1)
sinA(1−2sin
2
A)
=tanA(
cos2A
cos2A
) [∵
1−2sin
2
A=cos2A
2cos
2
A−1=cos2A
]
=tanA=RHS
Hence Proved.
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