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**Unformatted text preview: **for all t
under consideration, we have no additional restrictions on t. The identity 1 + tan2 (t) =
sec2 (t) is valid for all t under discussion, so we substitute sec(t) = x to get 1 + tan2 (t) =
√
π
x2 . We get tan(t) = ± x2 − 1, but since t lies in 0, π ∪ π , 32 , tan(t) ≥ 0, so we
2
√
x
choose tan(t) = √ 2 − 1. Since we found no additional restrictions on t, the equivalence
tan(arcsec(x)) = x2 − 1 holds on the domain of arcsec(x), (−∞, −1] ∪ [1, ∞).
π
(b) If we let t = arccsc(4x), then csc(t) = 4x for t in 0, π ∪ π , 32 . Then cos(arccsc(4x)) =
2
cos(t) and our objective is to express the latter in terms of x. Since cos(t) is deﬁned
everywhere, we have no additional restrictions on t. From csc(t) = 4x, we have sin(t) =
1
1
2
2
csc(t) = 4x which allows us to use the Pythagorean Identity, cos (t) + sin (t) = 1, which holds for all values of t. We get cos2 (t) + 12
4x 16x2 −1
16x2 = 1, or cos(t) = ± √ =± 16x2 −1
.
4|x| π
If t lies in 0, π , then cos(t) ≥ 0. Otherwis...

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