Question

\[\text{ If }\sin 2A = \lambda \sin 2B, \text{ prove that }\frac{\tan (A + B)}{\tan (A - B)} = \frac{\lambda + 1}{\lambda - 1}\]

 

Answer 1

Given:
sin 2A = λ sin 2B
\[\Rightarrow \frac{\sin2A}{\sin2B} = \lambda\]

\[\Rightarrow \frac{\sin2A + \sin2B}{\sin2A - \sin2B} = \frac{\lambda + 1}{\lambda - 1}\]

\[ \Rightarrow \frac{2\sin\left( \frac{2A + 2B}{2} \right)\cos\left( \frac{2A - 2B}{2} \right)}{2\sin\left( \frac{2A - 2B}{2} \right)\cos\left( \frac{2A + 2B}{2} \right)} = \frac{\lambda + 1}{\lambda - 1} \left[ \because \sin A + \sin B = 2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) and \sin A - \sin B = 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right) \right]\]

\[ \Rightarrow \frac{\sin\left( A + B \right)\cos\left( A - B \right)}{\sin\left( A - B \right)\cos\left( A + B \right)} = \frac{\lambda + 1}{\lambda - 1}\]

\[ \Rightarrow \tan\left( A + B \right)\cot\left( A - B \right)=\frac{\lambda + 1}{\lambda - 1}\]
\[\Rightarrow\frac{\tan\left( A + B \right)}{\tan\left( A - B \right)}=\frac{\lambda + 1}{\lambda - 1}\]
Hence proved.