This is an extension of a question that I managed to solve but couldn’t solve the other parts.
Question:
If $\dfrac {\sin(A-B)}{\sin(A+B)}=\dfrac {5}{7}$, prove that $\dfrac{\tan(A)}{\tan(B)}=6$. If $A+B=90^\circ$, then show that $\tan(A)=\sqrt {6}$ and find the value of $\sin(2A)$. Do not use the calculator for this question.
Attempt:
I managed to prove the first part:
$$ \frac{\sin(A-B)}{\sin(A+B)}=\frac{\sin(A)\cos(B)-\sin(B)\cos(A)}{\sin(A)\cos(B)+\sin(B)\cos(A)}=\frac{5}{7} $$ Cross multiplying, $$ 7\left(\sin(A)\cos(B)-\sin(B)\cos(A)\right)= 5\left(\sin(A)\cos(B)+\sin(B)\cos(A)\right) $$ that yields $$ 2\sin(A)\cos(B)=12\sin(B)\cos(A) $$ and thus $$\frac{\tan(A)}{\tan(B)}=6$$
But I’m stuck proving $\tan(A)=\sqrt{6}$ and finding $\sin(2A)$.
Please help. Thank you!