# If angle θ is divided into two parts such that the tangents of one part is λ times the tangent of other,

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 $\sin \theta=\frac{\lambda+1}{\lambda-1} \sin \phi$. [NCERT EXEMPLER]

Solution:

Let $\alpha$ and $\beta$ be the two parts of angle $\theta$. Then,

$\theta=\alpha+\beta$ and $\phi=\alpha-\beta$ (Given)

Now,

$\tan \alpha=\lambda \tan \beta$    (Given)

$\Rightarrow \frac{\tan \alpha}{\tan \beta}=\frac{\lambda}{1}$

Applying componendo and dividendo, we get

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

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

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

$\Rightarrow \frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=\frac{\lambda+1}{\lambda-1}$

$\Rightarrow \frac{\sin \theta}{\sin \phi}=\frac{\lambda+1}{\lambda-1} \quad(\theta=\alpha+\beta$ and $\phi=\alpha-\beta)$

$\Rightarrow \sin \theta=\frac{\lambda+1}{\lambda-1} \sin \phi$