Question

If angle \[\theta\]  is divided into two parts such that the tangents of one part is \[\lambda\] times the tangent of other, and \[\phi\] is their difference, then show that\[\sin\theta = \frac{\lambda + 1}{\lambda - 1}\sin\phi\]

 

Answer 1

Let \[\alpha\]  and \[\beta\] be the two parts of angle \[\theta\]. Then, \[\theta = \alpha + \beta\] and

\[\phi = \alpha + \beta\]               (Given)
Now,
\[\tan\alpha = \lambda \tan\beta \left( \text{ Given }\right)\]
\[ \Rightarrow \frac{\tan\alpha}{\tan\beta} = \frac{\lambda}{1}\]
Applying componendo and dividendo, we get

\[\frac{\tan\alpha + \tan\beta}{\tan\alpha - \tan\beta} = \frac{\lambda + 1}{\lambda - 1}\]

\[ \Rightarrow \frac{\frac{\sin\alpha}{\cos\alpha} + \frac{\sin\beta}{\cos\beta}}{\frac{\sin\alpha}{\cos\alpha} - \frac{\sin\beta}{\cos\beta}} = \frac{\lambda + 1}{\lambda - 1}\]

\[ \Rightarrow \frac{\frac{\sin\alpha \cos\beta + \cos\alpha \sin\beta}{\cos\alpha \cos\beta}}{\frac{\sin\alpha \cos\beta - \cos\alpha \sin\beta}{\cos\alpha \cos\beta}} = \frac{\lambda + 1}{\lambda - 1}\]

\[ \Rightarrow \frac{\sin\left( \alpha + \beta \right)}{\sin\left( \alpha - \beta \right)} = \frac{\lambda + 1}{\lambda - 1}\]

\[\Rightarrow \frac{\sin\theta}{\sin\phi} = \frac{\lambda + 1}{\lambda - 1} \left( \theta = \alpha + \beta\text{ and }\phi = \alpha - \beta \right)\]
\[ \Rightarrow \sin\theta = \frac{\lambda + 1}{\lambda - 1}\sin\phi\]