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$\alpha$ 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

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  • $\begingroup$ It is always a good idea to start by changing the tangent and cotangents into sine and cosines.Pythagorean identities can help too. $\endgroup$ – user242559 May 9 '16 at 22:47
  • $\begingroup$ numerators and denominators can be factorized $\endgroup$ – G Cab May 9 '16 at 22:55
  • $\begingroup$ How? I can't see it please show me. $\endgroup$ – user335850 May 9 '16 at 22:58
  • $\begingroup$ @pisquare e.g. : $ x^{\,6} + x^{\,4} - 2 = \left( {x - 1} \right)\left( {x + 1} \right)\left( {2x^{\,2} + x^{\,4} + 2} \right) $ and similarly for the other three terms, and the fractions gets much simplified. $\endgroup$ – G Cab May 9 '16 at 23:20
  • $\begingroup$ @G Cab thank, $1+1-2=0$, then used division of polynomial. As for the other I can't see any values to make the equation zero! If you do show me another example, but I doubt it. $\endgroup$ – user335850 May 10 '16 at 3:55
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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$$ {{\sin ^{\,2} \alpha - 1} \over {\sin ^{\,2} \alpha - 2}} $$

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