Question

Prove the following:

If sin 2A = λsin 2B then prove that ${\displaystyle \frac{\tan (\text{A}+\text{B})}{\tan (\text{A}-\text{B})}=\frac{\lambda +1}{\lambda -1}}$

Answer 1

sin 2A = λsin 2B

∴ ${\displaystyle \frac{\sin 2\text{A}}{\sin 2\text{B}}=\frac{\lambda }{1}}$

∴ ${\displaystyle \frac{\sin 2\text{A}+\sin 2\text{B}}{\sin 2\text{A}-\sin 2\text{B}}=\frac{\lambda +1}{\lambda -1}}$

∴ ${\displaystyle \frac{2\sin \left(\frac{2\text{A}+2\text{B}}{2}\right)\cdot \cos \left(\frac{2\text{A}-2\text{B}}{2}\right)}{2\cos \left(\frac{2\text{A}+2\text{B}}{2}\right)\cdot \sin \left(\frac{2\text{A}-2\text{B}}{2}\right)}=\frac{\lambda +1}{\lambda -1}}$

∴ ${\displaystyle \frac{\sin (\text{A}+\text{B})\cdot \cos (\text{A}-\text{B})}{\cos (\text{A}+\text{B})\cdot \sin (\text{A}-\text{B})}=\frac{\lambda +1}{\lambda -1}}$

∴ tan(A + B) · cot(A – B) = ${\displaystyle \frac{\lambda +1}{\lambda -1}}$

∴ ${\displaystyle \frac{\tan (\text{A}+\text{B})}{\tan (\text{A}-\text{B})}=\frac{\lambda +1}{\lambda -1}}$.