Question
$\text{If}\ \sin2\text{A}=\lambda\sin2\text{B},$ prove that:
$\frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}=\frac{\lambda+1}{\lambda-1}$
$\frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}=\frac{\lambda+1}{\lambda-1}$
✓
Answer
We have,
$\sin2\text{A}=\lambda\sin2\text{B}$
$\Rightarrow\ \lambda=\frac{\sin2\text{A}}{\sin2\text{B}}$
Now,
$\frac{\lambda+1}{\lambda-1}=\frac{\frac{\sin2\text{A}}{\sin2\text{B}}+1}{\frac{\sin2\text{A}}{\sin2\text{B}}-1}$
$=\ \frac{\frac{\sin2\text{A}+\sin2\text{B}}{\sin2\text{B}}}{\frac{\sin2\text{A}-\sin2\text{B}}{\sin2\text{B}}}$
$=\ \frac{\sin2\text{A}+\sin2\text{B}}{\sin2\text{A}-\sin2\text{B}}$
$=\ \frac{2\sin\Big(\frac{2\text{A}+2\text{B}}{2}\Big)\cos\Big(\frac{2\text{A}-2\text{B}}{2}\Big)}{2\sin\Big(\frac{2\text{A}-2\text{B}}{2}\Big)\cos\Big(\frac{2\text{A}+2\text{B}}{2}\Big)}$
$=\ \frac{\sin(\text{A+B})\cos(\text{A}-\text{B})}{\sin(\text{A}-\text{B})\cos(\text{A+B})}$
$=\ \frac{\sin(\text{A+B})\cos(\text{A}-\text{B})}{\cos(\text{A}+\text{B})\sin(\text{A}-\text{B})}$
$=\ \frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}$
$\therefore\ \frac{\lambda+1}{\lambda-1}=\frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}$
$\Rightarrow\ \frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}=\frac{\lambda+1}{\lambda-1}$ Hence proved.
$\sin2\text{A}=\lambda\sin2\text{B}$
$\Rightarrow\ \lambda=\frac{\sin2\text{A}}{\sin2\text{B}}$
Now,
$\frac{\lambda+1}{\lambda-1}=\frac{\frac{\sin2\text{A}}{\sin2\text{B}}+1}{\frac{\sin2\text{A}}{\sin2\text{B}}-1}$
$=\ \frac{\frac{\sin2\text{A}+\sin2\text{B}}{\sin2\text{B}}}{\frac{\sin2\text{A}-\sin2\text{B}}{\sin2\text{B}}}$
$=\ \frac{\sin2\text{A}+\sin2\text{B}}{\sin2\text{A}-\sin2\text{B}}$
$=\ \frac{2\sin\Big(\frac{2\text{A}+2\text{B}}{2}\Big)\cos\Big(\frac{2\text{A}-2\text{B}}{2}\Big)}{2\sin\Big(\frac{2\text{A}-2\text{B}}{2}\Big)\cos\Big(\frac{2\text{A}+2\text{B}}{2}\Big)}$
$=\ \frac{\sin(\text{A+B})\cos(\text{A}-\text{B})}{\sin(\text{A}-\text{B})\cos(\text{A+B})}$
$=\ \frac{\sin(\text{A+B})\cos(\text{A}-\text{B})}{\cos(\text{A}+\text{B})\sin(\text{A}-\text{B})}$
$=\ \frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}$
$\therefore\ \frac{\lambda+1}{\lambda-1}=\frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}$
$\Rightarrow\ \frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}=\frac{\lambda+1}{\lambda-1}$ Hence proved.
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