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True-or-False :If $\tan A=\large\frac{1-\cos B}{\sin B}$ then $\tan 2A=\tan B$

$\begin{array}{1 1}(A)\;\text{True}\\(B)\;\text{False}\end{array} $

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  • $\tan 2A=\large\frac{2\tan A}{1-\tan^2A}$
$\tan 2A=\large\frac{2\tan A}{1-\tan^2A}$
$\Rightarrow \large\frac{2(\Large\frac{1-\cos B}{\sin B}}{1-\Large\frac{(1-\cos B)^2}{\sin ^2B}}$
$\Rightarrow \large\frac{2(1-\cos B)}{\sin B}\times \frac{\sin ^2B}{\sin ^2B-(1-\cos B)^2}$
$\Rightarrow \large\frac{2\sin B(1-\cos B)}{\sin ^2B-(1+\cos ^2B-2\cos B)}$
$\Rightarrow \large\frac{2\sin B(1-\cos B)}{\sin ^2B-1-\cos ^2B+2\cos B}$
$\Rightarrow \large\frac{2\sin B(1-\cos B)}{-\cos ^2B-\cos ^2B+2\cos B}$
$\Rightarrow \large\frac{2\sin B(1-\cos B)}{-2\cos ^2B+2\cos B}$
$\Rightarrow \large\frac{2\sin B(1-\cos B)}{2\cos B(1-\cos B)}$
$\Rightarrow \tan B$
Hence the given statement is true.
answered May 20, 2014 by sreemathi.v

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