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Exercise 9(a) · Q6

Q.Express tanh⁡−1(12)\tanh^{-1}\left(\dfrac{1}{2}\right) in logarithmic form and hence evaluate it.

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Concept understanding — Logarithmic Forms of Inverse Hyperbolic Functions

Every inverse hyperbolic function can be rewritten as an explicit natural logarithm, and this is what makes the inverse functions computable rather than merely abstract symbols. The three principal theorems are:

sinh⁡−1x=log⁡ ⁣(x+x2+1) for all real x,cosh⁡−1x=log⁡ ⁣(x+x2−1) for x≥1,\sinh^{-1}x=\log\!\left(x+\sqrt{x^2+1}\right)\ \text{for all real }x,\qquad \cosh^{-1}x=\log\!\left(x+\sqrt{x^2-1}\right)\ \text{for }x\ge1,

tanh⁡−1x=12log⁡1+x1−x for −1<x<1.\tanh^{-1}x=\frac12\log\frac{1+x}{1-x}\ \text{for }-1<x<1.

The proof strategy is identical in every case: set yy equal to the inverse function, write xx in terms of eye^y and e−ye^{-y} using the original definition, and multiply through by eye^y (or e2ye^{2y}) to turn the equation into a genuine quadratic in eye^y. Solving that quadratic, exactly one root is admissible (the other is rejected because eye^y must be positive, or because it lies outside the principal branch), and taking a logarithm of the surviving root recovers yy. …

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