Assertion (A): coth x = frac{1-k}{1+k} where | vedclass

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(D) Assertion $(A)$: $\coth x = \frac{1-k}{1+k}$$\Rightarrow \frac{e^x + e^{-x}}{e^x - e^{-x}} = \frac{1-k}{1+k}$Applying Componendo and Dividendo:$\f

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$(A)$ is false but $(R)$ is true

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(D) Assertion $(A)$: $\coth x = \frac{1-k}{1+k}$$\Rightarrow \frac{e^x + e^{-x}}{e^x - e^{-x}} = \frac{1-k}{1+k}$Applying Componendo and Dividendo:$\frac{(e^x + e^{-x}) + (e^x - e^{-x})}{(e^x + e^{-x}) - (e^x - e^{-x})} = \frac{(1-k) + (1+k)}{(1-k) - (1+k)}$$\Rightarrow \frac{2e^x}{2e^{-x}} = \frac{2}{-2k}$$\Rightarrow e^{2x} = -\frac{1}{k}$Since $e^{2x} > 0$ for all $x \in \mathbb{R}$ and $-\frac{1}{k} Reason $(R)$: The function $y = 	anh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$. As $x 	o \infty$,$y 	o 1$ and as $x 	o -\infty$,$y 	o -1$. The graph lies strictly between $y = -1$ and $y = 1$. Thus,the reason is true.

Assertion $(A)$: $\coth x = \frac{1-k}{1+k}$ where $0 < k < 2$.
Reason $(R)$: The graph of $y = \tanh x$ always lies between the lines $y = -1$ and $y = 1$.
Choose the correct option:

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Assertion $(A)$: $\coth x = \frac{1-k}{1+k}$ where $0 < k < 2$.Reason $(R)$: The graph of $y = \tanh x$ always lies between the lines $y = -1$ and $y = 1$.Choose the correct option: