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If $\theta+\phi=\alpha$ and $\tan \theta=\mathrm{k} \tan \phi$ (where $\mathrm{K}>1)$, then the value of $\sin (\theta-\phi)$ is
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Verified Answer
The correct answer is:
$\left(\frac{\mathrm{k}-1}{\mathrm{k}+1}\right) \sin \alpha$
We have $\tan \theta=\mathrm{k} \tan \phi$ and $\theta+\phi=\alpha$
$$
\therefore \frac{\tan \theta}{\tan \phi}=\frac{k}{1}
$$
By Componendo Dividendo, we get
$$
\begin{aligned}
& \frac{\tan \theta+\tan \phi}{\tan \theta-\tan \phi}=\frac{\mathrm{k}+1}{\mathrm{k}-1} \\
& \therefore \frac{\frac{\sin \theta}{\cos \theta}+\frac{\sin \phi}{\cos \phi}}{\frac{\sin \theta}{\cos \theta}-\frac{\sin \phi}{\cos \phi}}=\frac{k+1}{k-1} \\
& \therefore \frac{\sin \cos \phi+\cos \theta \sin \phi}{\sin \theta \cos \phi-\cos \theta \sin \phi}=\frac{k+1}{k-1} \\
& \therefore \frac{\sin (\theta+\phi)}{\sin (\theta-\phi)}=\frac{\mathrm{k}+1}{\mathrm{k}-1} \Rightarrow \frac{\sin \alpha}{\sin (\theta-\phi)}=\frac{\mathrm{k}+1}{\mathrm{k}-1} \\
& \therefore \sin (\theta-\phi)=\frac{\mathrm{k}-1}{\mathrm{k}+1}(\sin \alpha) \\
&
\end{aligned}
$$
$$
\therefore \frac{\tan \theta}{\tan \phi}=\frac{k}{1}
$$
By Componendo Dividendo, we get
$$
\begin{aligned}
& \frac{\tan \theta+\tan \phi}{\tan \theta-\tan \phi}=\frac{\mathrm{k}+1}{\mathrm{k}-1} \\
& \therefore \frac{\frac{\sin \theta}{\cos \theta}+\frac{\sin \phi}{\cos \phi}}{\frac{\sin \theta}{\cos \theta}-\frac{\sin \phi}{\cos \phi}}=\frac{k+1}{k-1} \\
& \therefore \frac{\sin \cos \phi+\cos \theta \sin \phi}{\sin \theta \cos \phi-\cos \theta \sin \phi}=\frac{k+1}{k-1} \\
& \therefore \frac{\sin (\theta+\phi)}{\sin (\theta-\phi)}=\frac{\mathrm{k}+1}{\mathrm{k}-1} \Rightarrow \frac{\sin \alpha}{\sin (\theta-\phi)}=\frac{\mathrm{k}+1}{\mathrm{k}-1} \\
& \therefore \sin (\theta-\phi)=\frac{\mathrm{k}-1}{\mathrm{k}+1}(\sin \alpha) \\
&
\end{aligned}
$$
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