If 2A= 2B, then write the value of +1 -1 .

Question

If $\sin2\text{A}=\lambda\sin2\text{B},$ then write the value of $\frac{\lambda+1}{\lambda-1}.$

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Answer

We have, $\sin2\text{A}=\lambda\sin2\text{B}$ $\Rightarrow\ \frac{\sin2\text{A}}{\sin2\text{B}}=\lambda$ $\Rightarrow\ \frac{\sin2\text{A}}{\sin2\text{B}}+1=\lambda+1$ $\Rightarrow\ \frac{\sin2\text{A}+\sin2\text{B}}{\sin2\text{B}}=\lambda+1...(\text{i})$ Again, $\sin2\text{A}=\lambda\sin2\text{B}$ $\Rightarrow\ \frac{\sin2\text{A}}{\sin2\text{B}}=\lambda$ $\Rightarrow\ \frac{\sin2\text{A}}{\sin2\text{B}}-1=\lambda-1$ $\Rightarrow\ \frac{\sin2\text{A}-\sin2\text{B}}{\sin2\text{B}}=\lambda-1...(\text{ii})$ Dividing equation (i) by equation (ii), we get $\frac{\sin2\text{A}+\sin2\text{B}}{\sin2\text{A}-\sin2\text{B}}=\frac{\lambda+1}{\lambda-1}$ $\Rightarrow\ \frac{2\sin\Big(\frac{2\text{A}+2\text{B}}{2}\Big)\cos\Big(\frac{2\text{A}-2\text{B}}{2}\Big)}{2\sin\Big(\frac{2\text{A}-2\text{B}}{2}\Big)\cos\Big(\frac{2\text{A}+2\text{B}}{2}\Big)}=\frac{\lambda+1}{\lambda-1}$ $\Rightarrow\ \frac{\sin(\text{A+B})\cos(\text{A}-\text{B})}{\sin(\text{A}-\text{B})\cos(\text{A}+\text{B})}=\frac{\lambda+1}{\lambda-1}$ $\Rightarrow\ \frac{\sin(\text{A+B})\cos(\text{A}-\text{B})}{\cos(\text{A+B})\sin(\text{A}-\text{B})}=\frac{\lambda+1}{\lambda-1}$ $\Rightarrow\ \frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}=\frac{\lambda+1}{\lambda-1}$ $\therefore\ \frac{\lambda+1}{\lambda-1}=\frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}.$