If tan of one part is λ times the other, Show that sinθ = ((λ+1)/(λ−1)) sinϕ

Question

If angle \[ \theta \] is divided into two parts such that the tangent of one part is \[ \lambda \] times the tangent of the other, and \[ \phi \] is their difference, show that:

\[ \sin\theta = \frac{\lambda+1}{\lambda-1}\sin\phi \]

Proof

Let the two parts be \[ A \] and \[ B \] such that:

\[ A+B=\theta \]

and

\[ A-B=\phi \]

Given,

\[ \tan A=\lambda\tan B \]

Now,

\[ \sin\theta = \sin(A+B) \]

\[ = \sin A\cos B+\cos A\sin B \]

\[ = \cos A\cos B(\tan A+\tan B) \]

\[ = \cos A\cos B(\lambda\tan B+\tan B) \]

\[ = (\lambda+1)\cos A\sin B \]

Also,

\[ \sin\phi = \sin(A-B) \]

\[ = \sin A\cos B-\cos A\sin B \]

\[ = \cos A\cos B(\tan A-\tan B) \]

\[ = \cos A\cos B(\lambda\tan B-\tan B) \]

\[ = (\lambda-1)\cos A\sin B \]

Therefore,

\[ \frac{\sin\theta}{\sin\phi} = \frac{\lambda+1}{\lambda-1} \]

\[ \sin\theta = \frac{\lambda+1}{\lambda-1}\sin\phi \]

Hence proved.