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Question Number 74966 by ajfour last updated on 05/Dec/19

$$If\sin 2A=\lambda \sin 2B$$$$Provethat\frac{\tan (A+B)}{\tan (A-B)}=\frac{\lambda +1}{\lambda -1}.$$

Commented by Prithwish sen last updated on 05/Dec/19

$$\frac{\sin 2\mathrm{A}}{\sin 2\mathrm{B}}=\lambda$$$$\frac{\sin 2\mathrm{A}+\sin 2\mathrm{B}}{\sin 2\mathrm{A}-\sin 2\mathrm{B}}=\frac{\lambda +1}{\lambda -1}$$$$\frac{\sin (\mathrm{A}+\mathrm{B})\cos (\mathrm{A}-\mathrm{B})}{\cos (\mathrm{A}+\mathrm{B})\sin (\mathrm{A}-\mathrm{B})}=\frac{\lambda +1}{\lambda -1}$$$$\frac{\mathbf{tan}(\mathbf{A}+\mathbf{B})}{\mathbf{tam}(\mathbf{A}-\mathbf{B})}=\frac{\mathit{\lambda }-1}{\mathit{\lambda }+1}\mathbf{proved}.$$

Commented by ajfour last updated on 05/Dec/19

$$Thanks.$$

Answered by ajfour last updated on 05/Dec/19

$$\sin [(A+B)+(A-B)]=\lambda \sin [A+B-(A-B)]$$$$\Rightarrow$$$$\sin (A+B)\cos (A-B)+\cos (A+B)\sin (A-B)$$$$-\lambda \{\sin (A+B)\cos (A-B)-\cos (A+B)\sin (A-B)=0$$$$dividingby\cos (A+B)\cos (A-B)$$$$\tan (A+B)+\tan (A-B)$$$$=\lambda \{\tan (A+B)-\tan (A-B)$$$$\Rightarrow \frac{\tan (A+B)}{\tan (A-B)}=\frac{\lambda +1}{\lambda -1}.$$

Commented by Prithwish sen last updated on 05/Dec/19

$$\mathrm{Wow}!$$

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