Math, asked by AarohanNayakprogamer, 5 months ago

prove the above question

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Answered by madhulika7
1

Answer:

L.H.S=

 \frac{ \sinθ -  { 2sin }^{3}  θ}{2 { \cos}^{3}θ -  \cosθ  } \\   =   \frac{ \sinθ (1 - 2 { \sin }^{2} θ) }{ \cosθ(2 { \cos }^{2} θ - 1) }   \\  =  \frac{ \sinθ  (1 - 2(1 - 2 { \cos }^{2} θ))}{ \cosθ(2 { \cos}^{2}θ - 1) } \\ =  \frac{ \sinθ(2 { \cos }^{2}θ - 1) }{ \cosθ(2 { \cos }^{2} θ  - 1) }  \ \\  =  \frac{ \sin θ}{ \cos θ}  \\  =  \tan θ

= R.H.S

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