KnowledgeBoat Logo
|

Mathematics

Prove the following identities :

sin θ - 2 sin3θ2 cos3θ− cos θ\dfrac{\text{sin θ - 2 sin}^3 θ}{\text{2 cos}^3 θ - \text{ cos θ}} = tan θ

Trigonometric Identities

58 Likes

Answer

Solving L.H.S. of the equation :

⇒sin θ(1 - 2 sin2θ)cos θ(2 cos2θ−1)\Rightarrow \dfrac{\text{sin θ(1 - 2 sin}^2 θ)}{\text{cos θ(2 cos}^2 θ - 1)}

By formula,

sin2 θ = 1 - cos2 θ

⇒sin θ[1 - 2(1 - cos2θ)]cos θ(2 cos2θ−1)⇒tan θ×1−2+ 2 cos2θ(2 cos2θ−1)⇒tan θ×2 cos2θ−12 cos2θ−1⇒tan θ.\Rightarrow \dfrac{\text{sin θ[1 - 2(1 - cos}^2 θ)]}{\text{cos θ(2 cos}^2 θ - 1)} \\[1em] \Rightarrow \text{tan θ} \times \dfrac{1 - 2 + \text{ 2 cos}^2 θ}{\text{(2 cos}^2 θ - 1)} \\[1em] \Rightarrow \text{tan θ} \times \dfrac{\text{2 cos}^2 θ - 1}{\text{2 cos}^2 θ - 1} \\[1em] \Rightarrow \text{tan θ}.

Since, L.H.S. = R.H.S.

Hence, proved that sin θ - 2 sin3θ2 cos3θ− cos θ\dfrac{\text{sin θ - 2 sin}^3 θ}{\text{2 cos}^3 θ - \text{ cos θ}} = tan θ.

Answered By

22 Likes

Related Questions