$If \frac{\tan 3A}{\tan A}=a$ then $\frac{\sin 3A}{\sin A}=$

$\begin{array}{rcl}\frac{\tan 3\mathrm{A}}{\tan \mathrm{A}}& =& \mathrm{a}\\ \frac{1}{\tan \mathrm{A}}\left[\frac{3\tan \mathrm{A}-{\tan }^3\mathrm{A}}{1-3{\tan }^2\mathrm{A}}\right]& =& \mathrm{a}\\ \frac{3-{\tan }^2\mathrm{A}}{1-3{\tan }^2\mathrm{A}}& =& \mathrm{a}\\ \frac{3-{\tan }^2\mathrm{A}+1-3{\tan }^2\mathrm{a}}{3-{\tan }^2\mathrm{a}-1+3{\tan }^2\mathrm{a}}& =& \frac{\mathrm{a}+1}{\mathrm{a}-1}\\ \frac{4-4{\tan }^2\mathrm{A}}{2+2{\tan }^2\mathrm{A}}& =& \frac{\mathrm{a}+1}{\mathrm{a}-1}\\ 2\left(\frac{1-{\tan }^2\mathrm{a}}{1+{\tan }^2\mathrm{a}}\right)& =& \frac{\mathrm{a}+1}{\mathrm{a}-1}\\ 2\cos 2\mathrm{A}& =& \frac{\mathrm{a}+1}{(\mathrm{a}-1)}\end{array}$

                                  $\begin{array}{rcl}\frac{\sin 3A}{\sin \mathrm{A}}& =& \frac{3\sin \mathrm{A}-4{\sin }^3\mathrm{A}}{\sin \mathrm{A}}\\ & =& 3-4{\sin }^{2}A\\ & =& 3-2(1-\cos 2\mathrm{A})\\ & =& 3-2+2\cos 2\mathrm{A}\\ & =& 1+2\cos 2\mathrm{A}\\ & =& 1+\frac{\mathrm{a}+1}{\mathrm{a}-1}\\ & =& \frac{2\mathrm{a}}{\mathrm{a}-1}\end{array}$