Consider the following statements:Statement-I | vedclass

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(A) For Statement-$I$: $\operatorname{Cosh}^{-1} x = \ln(x + \sqrt{x^2 - 1})$ for $x \ge 1$. $\operatorname{Tanh}^{-1} x = \frac{1}{2} \ln(\frac{1+x}{

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Both statements $I$ and $II$ are true

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(A) For Statement-$I$: $\operatorname{Cosh}^{-1} x = \ln(x + \sqrt{x^2 - 1})$ for $x \ge 1$. $\operatorname{Tanh}^{-1} x = \frac{1}{2} \ln(\frac{1+x}{1-x})$ for $|x| For Statement-$II$: $\operatorname{Cosh}^{-1} x = \operatorname{Coth}^{-1} x$. $\operatorname{Coth}^{-1} x = \frac{1}{2} \ln(\frac{x+1}{x-1})$ for $|x| > 1$. Setting $\ln(x + \sqrt{x^2 - 1}) = \frac{1}{2} \ln(\frac{x+1}{x-1})$, we square both sides to get $(x + \sqrt{x^2 - 1})^2 = \frac{x+1}{x-1}$. Simplifying leads to $x^2 + x^2 - 1 + 2x\sqrt{x^2 - 1} = \frac{x+1}{x-1}$. Solving this equation yields one valid solution for $x > 1$. Thus, Statement-$II$ is true.

Consider the following statements:
Statement-$I$: $\operatorname{Cosh}^{-1} x = \operatorname{Tanh}^{-1} x$ has no solution.
Statement-$II$: $\operatorname{Cosh}^{-1} x = \operatorname{Coth}^{-1} x$ has only one solution.
The correct answer is:

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Consider the following statements:Statement-$I$: $\operatorname{Cosh}^{-1} x = \operatorname{Tanh}^{-1} x$ has no solution.Statement-$II$: $\operatorname{Cosh}^{-1} x = \operatorname{Coth}^{-1} x$ has only one solution.The correct answer is: