Question

Solve the following:

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

Answer 1

consider the problem

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

$\begin{array}{c}=\frac{\sin \theta \left(1-2{\sin }^2\theta \right)}{\cos \theta \left(2{\cos }^2\theta -1\right)}\\ =\frac{\sin \theta \left(1-2\left(1-{\cos }^2\theta \right)\right)}{\cos \theta \left(2{\cos }^2\theta -1\right)}\left(\sin ce{\sin }^2\theta +{\cos }^2\theta =1\right)\\ =\frac{\sin \theta \left(2{\cos }^2\theta -1\right)}{\cos \theta \left(2{\cos }^2\theta -1\right)}\\ =\frac{\sin \theta }{\cos \theta }\\ =\tan \theta \end{array}$