**Unformatted text preview: **o π , π then tan(t) ≤ 0. As a
2
result, we get a piecewise deﬁned function for tan(t)
√ tan(t) = x2 − 1, if 0 ≤ t < π
2
√
2 − 1, if π < t ≤ π
−x
2 Now we need to determine what these conditions on t mean for x. We know that the
domain of arcsec(x) is (−∞, −1] ∪ [1, ∞), and since x = sec(t), x ≥ 1 corresponds to
0 ≤ t < π , and x ≤ −1 corresponds to π < t ≤ π . Since we encountered no further
2
2
restrictions on t, the equivalence below holds for all x in (−∞, −1] ∪ [1, ∞).
√ tan(arcsec(x)) = x2 − 1,
if x ≥ 1
√
− x2 − 1, if x ≤ −1 (b) To simplify cos(arccsc(4x)), we start by letting t = arccsc(4x). Then csc(t) = 4x for
t in π , 0 ∪ 0, π . Our objective is to write cos(arccsc(4x)) = cos(t) in terms of
2
2
x. Since cos(t) is deﬁned for all t, we do not encounter any additional restrictions on
t. From csc(t) = 4x, we get sin(t) = 41 . The identity cos2 (t) + sin2 (t) = 1 holds
x
2
for all values of t and substituting for sin(t) yields...

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