Answer
See proof
Work Step by Step
$\frac{\tan\frac{5t}{2}}{\tan\frac{t}{2}}$
$=\frac{\frac{\sin\frac{5t}{2}}{\cos\frac{5t}{2}}}{\frac{\sin\frac{t}{2}}{\cos\frac{t}{2}}}$
$=\frac{\sin\frac{5t}{2}\cos\frac{t}{2}}{\sin\frac{t}{2}\cos\frac{5t}2{}}$
$=\frac{\frac{1}{2}(\sin(\frac{5t}{2}+\frac{t}{2})+\sin(\frac{5t}{2}-\frac{t}{2}))}{\frac{1}{2}(\sin(\frac{t}{2}+\frac{5t}{2})+\sin(\frac{t}{2}-\frac{5t}{2}))}$
$=\frac{\sin3t+\sin2t}{\sin3t+\sin(-2t)}$
$=\frac{\sin3t+\sin2t}{\sin3t-\sin2t}$
