If 2A= 2B, prove that: (A+B) (A-B) = +1 -1 - Vidyadip

Question

$\text{If}\ \sin2\text{A}=\lambda\sin2\text{B},$ prove that: $\frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}=\frac{\lambda+1}{\lambda-1}$

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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.

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