Question

Prove that
sin thita-2 sine^3thita/
2 cos^3thita-cos thita =
tan thita​

Answer 1

Answer:

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