<s>[INST] Trigonometric identities involved $\sin(\alpha)$, $\cos(\alpha)$, $\tan(\alpha)$ $\alpha$ is valid between  $-90<\alpha<90$ Degrees
Show that,
$$ \frac{\tan^6(\alpha)-\tan^4(\alpha)+2}{\tan^6(\alpha)-2\tan^2(\alpha)+4 }\cdot\cos^2(\alpha) =
  \frac{\sin^6(\alpha)+\sin^4(\alpha)-2}{\sin^6(\alpha)-2\sin^2(\alpha)-4 } $$

$$
  \frac{\sin^6(\alpha)+\sin^4(\alpha)-2}{\sin^6(\alpha)-2\sin^2(\alpha)-4 } =\frac{(\sin^2(\alpha)-1)(\sin^4(\alpha)+2\sin(\alpha)+2)} {(\sin^2(\alpha)-4)(\sin^4(\alpha)+2\sin(\alpha)+2)} $$
$$
  \frac{\sin^6(\alpha)+\sin^4(\alpha)-2}{\sin^6(\alpha)-2\sin^2(\alpha)-4 } =\frac{\sin^2(\alpha)-1} {\sin^2(\alpha)-4} $$
Can this expression be simplify more further?
$$ \frac{\tan^6(\alpha)-\tan^4(\alpha)+2}{\tan^6(\alpha)-2\tan^2(\alpha)+4 }\cdot\cos^2(\alpha)$$
please give a hand here can't seem to do it
 [/INST] So, for the LH fraction:
$$
{{x^{\,6}  - x^{\,4}  + 2} \over {x^{\,6}  - 2x^{\,2}  + 4}} = {{y^{\,3}  - y^{\,2}  + 2} \over {y^{\,3}  - 2y + 4}}
$$
then by Ruffini's method
$$
 = {{\left( {y + 1} \right)\left( {y^{\,2}  - 2y^{\,2}  + 1} \right)} \over {\left( {y + 2} \right)\left( {y^{\,2}  - 2y^{\,2}  + 1} \right)}} = {{x^{\,2}  + 1} \over {x^{\,2}  + 2}}
$$
Same method for the RH fraction, as already done, but corrected to$$
{{\sin ^{\,2} \alpha  - 1} \over {\sin ^{\,2} \alpha  - 2}}
$$
</s>