If $ heta+\phi=\alpha$ and $ an heta=\mathrm{k} an \phi$ (where $\mathrm{K}1)$, then the value of $\sin ( heta-\phi)$ is

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If $	heta+\phi=\alpha$ and $	an 	heta=\mathrm{k} 	an \phi$ (where $\mathrm{K}>1)$, then the value of $\sin (	heta-\phi)$ is

Aicharya · Accepted Answer

We have $	an 	heta=\mathrm{k} 	an \phi$ and $	heta+\phi=\alpha$ $	herefore \frac{	an 	heta}{	an \phi}=\frac{k}{1}$ By Componendo Dividendo, we get $\begin{aligned} & \frac{	an 	heta+	an \phi}{	an 	heta-	an \phi}=\frac{\mathrm{k}+1}{\mathrm{k}-1} \ & 	herefore \frac{\frac{\sin 	heta}{\cos 	heta}+\frac{\sin \phi}{\cos \phi}}{\frac{\sin 	heta}{\cos 	heta}-\frac{\sin \phi}{\cos \phi}}=\frac{k+1}{k-1} \ & 	herefore \frac{\sin \cos \phi+\cos 	heta \sin \phi}{\sin 	heta \cos \phi-\cos 	heta \sin \phi}=\frac{k+1}{k-1} \ & 	herefore \frac{\sin (	heta+\phi)}{\sin (	heta-\phi)}=\frac{\mathrm{k}+1}{\mathrm{k}-1} \Rightarrow \frac{\sin \alpha}{\sin (	heta-\phi)}=\frac{\mathrm{k}+1}{\mathrm{k}-1} \ & 	herefore \sin (	heta-\phi)=\frac{\mathrm{k}-1}{\mathrm{k}+1}(\sin \alpha) \ & \end{aligned}$

If $\theta+\phi=\alpha$ and $\tan \theta=\mathrm{k} \tan \phi$ (where $\mathrm{K}>1)$, then the value of…

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