Question

Prove that:

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

Answer 1

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

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

$=\frac{sin\theta }{cos\theta }\left[\frac{1-2si{n}^2\theta }{2(1-si{n}^2\theta )-1}\right]$

$=tan\theta \left[\frac{1-2si{n}^2\theta }{2-2si{n}^2\theta -1}\right]$

$=tan\theta \left[\frac{1-2si{n}^2\theta }{1-2si{n}^2\theta }\right]$

$=tan\theta =RHS$