upwards acceleration and friction

assuming that the statement 2F_gx=m(a⁺+a^-), where F_gx is the horizontal component of the force of gravity on an inclined plane θ, m is the mass of the object on the plane, and a⁺ and a^- are the accelerations uphill and downhill with respect to the plane respectively, the following formulae may be derived:

$\theta=\arcsin\frac{a^++a^-}{2g}$
$a^+=g(\sin\theta+\mu_k\cos\theta)$
$a^-=g(\sin\theta-\mu_k\cos\theta)$
$\mu_k=\frac{a^+\sec\theta}{g}-\tan\theta$
$\mu_k=-\frac{a^-\sec\theta}{g}+\tan\theta$
$a^++a^-=2g\sin\theta$
$a^+-a^-=2\mu_k\cos\theta$
$\mu_k=\frac{\sec\theta(a^+-a^-)}{2}$
$\mu_k=\pm\frac{a^{\pm}\sec\theta}{g}\mp\tan\theta$

which are algebraically derived from the following:

$2F_{g_x}=m(a^++a^-)$
$F_{g_x}=\frac{m(a^++a^-)}{2}$
$mg_x=\frac{m(a^++a^-)}{2}$
$mg\sin\theta=\frac{m(a^++a^-)}{2}$
$g\sin\theta=\frac{a^++a^-}{2}$
$\sin\theta=\frac{a^++a^-}{2g}$
$\underline{\theta=\arcsin{\frac{a^++a^-}{2}}}$
$\Box$

$\underline{\mathrm{note:}}\begin{array}{l} \Sigma{\vec{F}^+}=\vec{F}_{g_x}+\vec{F}_{\mu} \\ \Sigma{\vec{F}^-}=\vec{F}_{g_x}-\vec{F}_{\mu} \\ F_{\mu}=F_N\mu_k \\ F_{\mu}=mg\mu_k\cos\theta\end{array}\therefore$
$ma^+=mg\sin\theta+mg\mu_k\cos\theta\quad,\quad ma^-=mg\sin\theta-mg\mu_k\cos\theta$
$ma^+=mg(\sin\theta+\mu_k\cos\theta)\quad,\quad ma^-=mg(\sin\theta-\mu_k\cos\theta)$
$\underline{a^+=g(\sin\theta+\mu_k\cos\theta)}\quad,\quad\underline{a^-=g(\sin\theta-\mu_k\cos\theta)}$
$\Box$

$\frac{a^+}{g}=\sin\theta+\mu_k\cos\theta\quad,\quad\frac{a^-}{g}=\sin\theta-\mu_k\cos\theta$
$\frac{a^+}{g}-\sin\theta=\mu_k\cos\theta\quad,\quad\frac{a^-}{g}-\sin\theta=-\mu_k\cos\theta$
$\frac{a^+}{g\cos\theta}-\frac{\sin\theta}{\cos\theta}=\mu_k\quad,\quad-\frac{a^-}{g\cos\theta}+\frac{\sin\theta}{\cos\theta}=\mu_k$
$\underline{\frac{a^+\sec\theta}{g}-\tan\theta=\mu_k}\quad,\quad\underline{-\frac{a^-\sec\theta}{g}+\tan\theta=\mu_k}$
$\Box$

$\frac{a^+\sec\theta}{g}-\tan\theta=\mu_k=-\frac{a^-\sec\theta}{g}+\tan\theta$
$\frac{a^+\sec\theta}{g}-\tan\theta=-\frac{a^-\sec\theta}{g}+\tan\theta$
$a^+\sec\theta-g\tan\theta=-a^-\sec\theta+g\tan\theta$
$a^+-\frac{g\tan\theta}{\sec\theta}=-a^-+\frac{g\tan\theta}{\sec\theta}$
$a^+-g\sin\theta=-a^-+g\sin\theta$
$a^+=-a^-+2g\sin\theta$
$\underline{a^++a^-=2g\sin\theta}$
$\Box$

$a^+-a^-=g(\sin\theta+\mu_k\cos\theta)-g(\sin\theta-\mu_k\cos\theta)$
$a^+-a^-=g\sin\theta+\mu_k\cos\theta-g\sin\theta+\mu_k\cos\theta$
$\underline{a^+-a^-=2\mu_k\cos\theta}$
$\Box$

$2\mu_k\cos\theta=a^+-a^-$
$\mu_k\cos\theta=\frac{a^+-a^-}{2}$
$\mu_k=\frac{a^+-a^-}{2\cos\theta}$
$\underline{\mu_k=\frac{\sec\theta(a^+-a^-)}{2}}$
$\Box$