Math, asked by whatever49, 6 months ago

sinAcos³A-sin³A/2cos³A-cosA = tanA​

Answers

Answered by rajeevr06
0

Answer:

LHS,

 \frac{sinx( {cos}^{2} x -  {sin}^{2}x) }{cosx(2 {cos}^{2}x - 1) }  =  \frac{sinx \times cos2x}{cosx \times cos2x}  =  \tan(x)

RHS

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