Question

$provethat\frac{sin\theta -2s{in}^3\theta }{2co{s}^3\theta -cos\theta }=tan\theta$

Answer 1

LHS

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

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

$=tan\theta \frac{[1+2(1-co{s}^2\theta \left)\right]}{2co{s}^2\theta -1}$

$=tan\theta \frac{2co{s}^2\theta -1}{2co{s}^2\theta -1}$

$=tan\theta$ = RHS