Question

please don't ans the question if u don't know plz plz plz .......​

Answer 1

Answer:

$\frac{sinθ - 2 {sin}^{3} θ}{2 {cos}^{3} θ- cosθ } = tanθ \\ \frac{sinθ(1 - 2 {sin}^{2}θ) }{cosθ(2 {cos}^{2} θ - 1)} = tanθ \\ \frac{sinθ(1 - {sin}^{2}θ + {sin}^{2} θ) }{cosθ( {cos}^{2}θ + {cos}^{2} θ - 1)} = tan θ\\ \frac{sinθ( {cos}^{2}θ + {sin}^{2}θ) }{cosθ({cos}^{2} θ+ {sin}^{2} θ)} = tan θ\\ \frac{sinθ(1)}{cosθ(1)} = tan θ\\ \frac{sinθ}{cosθ} = tanθ \\ tanθ = tanθ \\ LHS=RHS \\ </p><p>Hence \: proved$