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yields arcsec sec 54
= arcsec(− 2) = arccos − 22 = 34 . (d) One way to begin to simplify cot (arccsc (−3)) is to let t = arccsc(−3). Then, csc(t) = −3
and, since this is negative, we have that t lies in the interval − π , 0 . We are after
2
cot (arccsc (−3)) = cot(t), so we use the Pythagorean √
Identity √ + cot2 (t) = csc2 (t).
1
Substituting, we have 1 + cot2 (t) = (−3)2 , or cot(t) = ± 8 = ±2 2. Since − π ≤ t < 0,
2
√
cot(t) < 0, so we get cot (arccsc (−3)) = −2 2.
2. (a) We begin simplifying tan(arcsec(x)) by letting t = arcsec(x). Then, sec(t) = x for
t in 0, π ∪ π , π , and we seek a formula for tan(t). Since tan(t) is deﬁned for all
2
2
the t values in 0, π ∪ π , π , we have no additional restrictions on t. The identity
2
2
1 + tan2 (t) = sec2 (t) is valid for all values t under consideration, and we get substitute
√
sec(t) = x to get 1 + tan2 (t) = x2 . Hence, tan(t) = ± x2 − 1. If t belongs to 0, π
2
then tan(t) ≥ 0; if, on the the other hand, t belongs t...

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