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If $\cos A=m\cos B$ and $\cot\large\frac{A+B}{2}$$=\lambda\tan\large\frac{B-A}{2}$ then $\lambda$ is

$\begin{array}{1 1}(a)\;\large\frac{m}{m-1}&(b)\;\large\frac{m+1}{m}\\(c)\;\large\frac{m+1}{m-1}&(d)\;None\; of \;these\end{array}$

1 Answer

We have $\cos A=m\cos B$
$\Rightarrow \large\frac{\cos A}{\cos B}=\frac{m}{1}$
$\Rightarrow \large\frac{\cos A+\cos B}{\cos A-\cos B}=\frac{m+1}{m-1}$
$\Rightarrow \large\frac{2\cos\Large\frac{A+B}{2}\cos\Large\frac{B-A}{2}}{2\sin\Large\frac{A+B}{2}\sin\Large\frac{B-A}{2}}=\large\frac{m+1}{m-1}$
$\cot\large\frac{A+B}{2}=\big(\large\frac{m+1}{m-1}\big)$$\tan\large\frac{B-A}{2}$
$\Rightarrow \lambda=\large\frac{m+1}{m-1}$
Hence (c) is the correct answer.
answered Oct 15, 2013 by sreemathi.v
 

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