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If $\cos A=m \cos B$ and $\cot \left(\frac{A+B}{2}\right)=\lambda \tan \left(\frac{B-A}{2}\right)$, then $\lambda$ is equal to
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The correct answer is:
$\frac{m+1}{m-1}$
Given, $\cos A=m \cos B \Rightarrow \frac{\cos A}{\cos B}=m$
On applying componendo and dividendo rule, we get
$\begin{array}{ll}
& \frac{\cos A+\cos B}{\cos A-\cos B}=\frac{m+1}{m-1} \\
\Rightarrow \quad & \frac{2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{B-A}{2}\right)}=\frac{m+1}{m-1} \\
\Rightarrow \quad & \frac{\cot \left(\frac{A+B}{2}\right)}{\tan \left(\frac{B-A}{2}\right)}=\frac{m+1}{m-1} \\
\Rightarrow \quad & \cot \left(\frac{A+B}{2}\right)=\left(\frac{m+1}{m-1}\right) \tan \left(\frac{B-A}{2}\right) \\
\therefore \quad & \lambda=\frac{m+1}{m-1}
\end{array}$
On applying componendo and dividendo rule, we get
$\begin{array}{ll}
& \frac{\cos A+\cos B}{\cos A-\cos B}=\frac{m+1}{m-1} \\
\Rightarrow \quad & \frac{2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{B-A}{2}\right)}=\frac{m+1}{m-1} \\
\Rightarrow \quad & \frac{\cot \left(\frac{A+B}{2}\right)}{\tan \left(\frac{B-A}{2}\right)}=\frac{m+1}{m-1} \\
\Rightarrow \quad & \cot \left(\frac{A+B}{2}\right)=\left(\frac{m+1}{m-1}\right) \tan \left(\frac{B-A}{2}\right) \\
\therefore \quad & \lambda=\frac{m+1}{m-1}
\end{array}$
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