Question

An angle $\alpha$ is divided into two parts so that the ratio of the tangents of these parts is $\lambda$. If the difference between these parts is $x$ than $\frac{\sin x}{\sin \alpha }$ is equal to

• A. $\frac{\lambda }{(\lambda +1)}$
• B. $\frac{(\lambda -1)}{\lambda }$
• C. $\frac{\lambda -1}{\lambda +1}$
• D. none of these

Answer 1

The correct option is C $\frac{\lambda -1}{\lambda +1}$

Let ${\theta }_1+{\theta }_2=\alpha$ and ${\theta }_1-{\theta }_2=x$
Then, $\frac{tan{\theta }_1}{tan{\theta }_2}=\lambda$
Applying componendo and dividendo
$\frac{tan{\theta }_1+tan{\theta }_2}{tan{\theta }_1-tan{\theta }_2}=\frac{\lambda +1}{\lambda -1}\Rightarrow \frac{\sin {\theta }_1\cos {\theta }_2+\cos {\theta }_1\sin {\theta }_2}{\sin {\theta }_1\cos {\theta }_2+\cos {\theta }_1\sin {\theta }_2}=\frac{\lambda +1}{\lambda -1}\Rightarrow \frac{\sin ({\theta }_1+{\theta }_2)}{\sin ({\theta }_1-{\theta }_2)}=\frac{\lambda +1}{\lambda -1}\Rightarrow \frac{\sin \alpha }{\sin x}=\frac{\lambda +1}{\lambda -1}\Rightarrow \frac{\sin x}{\sin \alpha }=\frac{\lambda -1}{\lambda +1}$