$\frac{\tanh x}{sech x - 1} + \frac{\tanh x}{sech x + 1} =$

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$\cos e c h x$

$2 \cos e c h x$

$- \cos e c h x$

$- 2 \cos e c h x$

answer is D.

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Detailed Solution

$G i v e n \frac{\tan h x}{s e c h x - 1} + \frac{\tan h x}{s e c h x + 1} = = \frac{\frac{\sinh x}{\cosh x}}{\frac{1}{\cosh x} - 1} + \frac{\frac{\sinh x}{\cosh x}}{\frac{1}{\cosh x} + 1} = \frac{\frac{\sinh x}{\cosh x}}{\frac{1 - \cos h x}{\cos h x}} + \frac{\frac{\sinh x}{\cosh x}}{\frac{1 + \cos h x}{\cos h x}} = \frac{\sin h x}{1 - \cos h x} + \frac{\sin h x}{1 + \cos h x} = \sin h x (\frac{1}{1 - \cos h x} + \frac{1}{1 + \cos h x}) = \sin h x (\frac{2}{(1 - \cos h x) (1 + \cos h x)}) = \frac{2 \sin h x}{1 - \cos^{2} h^{} x} = \frac{2 \sin h x}{- \sin h^{2} x} = - 2 \cos e c h x$