Answer: (a) \(\tanh^2 t+\operatorname{sech}^2 t=1\).

(b) The length of \(C\) is \(1\).

(a) From the exponential definitions,

\(\tanh t=\dfrac{e^t-e^{-t}}{e^t+e^{-t}}, \qquad \operatorname{sech} t=\dfrac{2}{e^t+e^{-t}}.\)

Hence

\(\tanh^2 t+\operatorname{sech}^2 t=\left(\dfrac{e^t-e^{-t}}{e^t+e^{-t}}\right)^2+\dfrac{4}{(e^t+e^{-t})^2}\)

\(=\dfrac{(e^t-e^{-t})^2+4}{(e^t+e^{-t})^2}\)

\(=\dfrac{e^{2t}-2+e^{-2t}+4}{(e^t+e^{-t})^2}\)

\(=\dfrac{e^{2t}+2+e^{-2t}}{(e^t+e^{-t})^2}.\)

But

\((e^t+e^{-t})^2=e^{2t}+2+e^{-2t},\)

so the fraction is equal to \(1\). Therefore

\(\tanh^2 t+\operatorname{sech}^2 t=1.\)

(b) For the parametric curve,

\(x=\ln(\cosh t), \qquad y=\tan^{-1}(\sinh t), \qquad 0\le t\le 1.\)

Differentiate:

\(\dfrac{dx}{dt}=\dfrac{1}{\cosh t}\cdot \sinh t=\tanh t.\)

Also

\(\dfrac{dy}{dt}=\dfrac{1}{1+\sinh^2 t}\cdot \cosh t=\dfrac{\cosh t}{1+\sinh^2 t}.\)

Using \(1+\sinh^2 t=\cosh^2 t\),

\(\dfrac{dy}{dt}=\dfrac{\cosh t}{\cosh^2 t}=\dfrac{1}{\cosh t}=\operatorname{sech} t.\)

The arc length is

\(s=\int_0^1 \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt\)

\(=\int_0^1 \sqrt{\tanh^2 t+\operatorname{sech}^2 t}\,dt.\)

From part (a), \(\tanh^2 t+\operatorname{sech}^2 t=1\), so

\(s=\int_0^1 1\,dt=1.\)

Hence the length of \(C\) is \(1\).