Starting from the definitions of tanh and sech in terms of exponentials, prove that
$$1 - \tanh ^ { 2 } \theta = \operatorname { sech } ^ { 2 } \theta$$
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The variables \(x\) and \(y\) are such that \(\tanh y = \cos \left( x + \frac { 1 } { 4 } \pi \right)\), for \(- \frac { 1 } { 4 } \pi < x < \frac { 3 } { 4 } \pi\).
By differentiating the equation \(\tanh y = \cos \left( x + \frac { 1 } { 4 } \pi \right)\) with respect to \(x\), show that
$$\frac { \mathrm { dy } } { \mathrm { dx } } = - \operatorname { cosec } \left( \mathrm { x } + \frac { 1 } { 4 } \pi \right)$$
Hence find the first three terms in the Maclaurin's series for \(\tanh ^ { - 1 } \left( \cos \left( x + \frac { 1 } { 4 } \pi \right) \right)\) in the form \(\frac { 1 } { 2 } \ln a + b x + c x ^ { 2 }\), giving the exact values of the constants \(a , b\) and \(c\).