Prove that (sin θ – 2 sin^3 θ) / (2 cos^3 θ – cos θ) = tan θ

Let's start from LHS:LHS = \frac {\sin \theta – 2 \sin^3 \theta}{2 \cos ^3 \theta – \cos \theta}= \frac {\sin \theta (1 – 2 \sin^2 \theta)}{\cos \theta (2\cos ^2 \theta – 1)}We know that sin^2 θ + cos^2 θ = 1By substituting this value in the above equation, we get:LHS = \frac {\sin \theta ((\sin^2 \theta + \cos^2 \theta) – 2 \sin^2 \theta)}{\cos \theta (2\cos ^2 \theta – (\sin^2 \theta + \cos^2 \theta))}

Prove that (sin θ – 2 sin3 θ) / (2 cos3 θ

Q) Prove that $\frac {\sin \theta - 2 \sin^3 \theta}{2 \cos ^3 \theta - \cos \theta} = \tan \theta$

Ans: Let’s start from LHS:

LHS = $\frac {\sin \theta - 2 \sin^3 \theta}{2 \cos ^3 \theta - \cos \theta}$

= $\frac {\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (2\cos ^2 \theta - 1)}$

We know that sin² θ + cos² θ = 1

By substituting this value in the above equation, we get:

LHS = $\frac {\sin \theta ((\sin^2 \theta + \cos^2 \theta) - 2 \sin^2 \theta)}{\cos \theta (2\cos ^2 \theta - (\sin^2 \theta + \cos^2 \theta))}$

= $\frac {\sin \theta (\sin^2 \theta + \cos^2 \theta - 2 \sin^2 \theta)}{\cos \theta (2\cos ^2 \theta - \sin^2 \theta - \cos^2 \theta)}$

= $\frac {\sin \theta (\cancel{\sin^2 \theta} + \cos^2 \theta - \cancel{2} \sin^2 \theta)}{\cos \theta (\cancel{2}\cos ^2 \theta - \sin^2 \theta - \cancel{\cos^2 \theta)}}$