Math, asked by whatever49, 10 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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