If theta+phi=alpha and tan theta=k tan phi (w | vedclass

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(C) Given $	an 	heta = k 	an \phi$ and $	heta + \phi = \alpha$. $\frac{	an 	heta}{	an \phi} = \frac{k}{1}$ Applying Componendo and Dividendo: $

Vedclass · Accepted Answer

$\left(\frac{k-1}{k+1}\right) \sin \alpha$

Vedclass · Answer

(C) Given $	an 	heta = k 	an \phi$ and $	heta + \phi = \alpha$. $\frac{	an 	heta}{	an \phi} = \frac{k}{1}$ Applying Componendo and Dividendo: $\frac{	an 	heta + 	an \phi}{	an 	heta - 	an \phi} = \frac{k+1}{k-1}$ Converting to sine and cosine: $\frac{\frac{\sin 	heta}{\cos 	heta} + \frac{\sin \phi}{\cos \phi}}{\frac{\sin 	heta}{\cos 	heta} - \frac{\sin \phi}{\cos \phi}} = \frac{k+1}{k-1}$ $\frac{\sin 	heta \cos \phi + \cos 	heta \sin \phi}{\sin 	heta \cos \phi - \cos 	heta \sin \phi} = \frac{k+1}{k-1}$ Using the identity $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$: $\frac{\sin(	heta + \phi)}{\sin(	heta - \phi)} = \frac{k+1}{k-1}$ Substituting $	heta + \phi = \alpha$: $\frac{\sin \alpha}{\sin(	heta - \phi)} = \frac{k+1}{k-1}$ Therefore,$\sin(	heta - \phi) = \left(\frac{k-1}{k+1}ight) \sin \alpha$.

If $\theta+\phi=\alpha$ and $\tan \theta=k \tan \phi$ (where $k>1$),then the value of $\sin (\theta-\phi)$ is

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