Question

sin theta _ 2 sin3 theta / 2cos3 theta _ cos theta = tan theta

Answer 1

$\text{consider,}$

$\displaystyle\frac{sin\theta-2\,sin^3\theta}{2\,cos^3\theta-cos\theta}$

$=\displaystyle\frac{sin\theta[1-2\,sin^2\theta]}{cos\theta[2\,cos^2\theta-1]}$

$\text{using}$

$\boxed{\begin{minipage}{4cm} \bf\,cos2A=1-2\,sin^2A\\\\cos2A=2\,cos^2A-1 \end{minipage}}$

$=\displaystyle\frac{sin\theta{\times}cos2\theta}{cos\theta{\times}cos2\theta}$

$=\displaystyle\frac{sin\theta}{cos\theta}$

$=tan\theta$

$\implies\boxed{\bf\frac{sin\theta-2sin^3\theta}{2cos^3\theta-cos\theta}=tan\theta}$