Prove that (sinA-2sin³A)÷2cos³A-cosA=tanA
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Hola‼.......❤✌
LHS,
sin A ( 1 - sin ² A ) / [ cos A ( 2 cos ² A - 1 ) ]
sin A ( 2 cos ² A - 1 )
-------------------------------
cos A ( 2 cos ² A - 1 )
sin A
--------
cos A
tan A.
or,
sinA-2sin^3A/2cos^3A-cosA
= sinA [ 1 -2sin^2A] / cos A [ 2cos^2A- 1]
= sinA [ cos 2A] / cos A [cos 2A]
= sinA / cos A
= tan A
= R.H.S
Hope it’s helpful..... ☺
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