Prove the following identities , where the angles involved are acute angles for which the expression are defined.
$(i)\frac{sin\theta -2si{n}^3\theta }{2co{s}^3\theta -cos\theta }=tan\theta$

$LHS=\frac{sin\theta -2si{n}^3\theta }{2co{s}^3\theta -cos\theta }$

$=\frac{sin\theta (1-2si{n}^2\theta )}{cos\theta (2co{s}^2\theta -1)}$

$=\frac{sin\theta \times \left\{co{s}^2\theta +si{n}^2\theta -2si{n}^2\theta \right\}}{cos\theta \times \left\{2co{s}^2\theta -(co{s}^2\theta +si{n}^2\theta )\right\}}$

$=\frac{}{sin\theta \times (co{s}^2\theta -si{n}^2\theta )}cos\theta \times (co{s}^2\theta -si{n}^2\theta )$

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

$\therefore LHS=RHS$