Answer
$\displaystyle \sec^{2}t\cdot\sin^{2}t=\frac{1-\cos^{2}t}{\cos^{2}t}$
Work Step by Step
see: p.415, Reciprocal Identities:
$\displaystyle \csc t=\frac{1}{\sin t} \qquad \displaystyle \sec t=\frac{1}{\cos t}\qquad \displaystyle \cot t=\frac{1}{\tan t}$
and: Pythagorean Identities:
$\sin^{2}t+\cos^{2}t=1\qquad\tan^{2}t+1=\sec^{2}t\qquad 1 +\cot^{2}t=\csc^{2}t$
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$\displaystyle \sec t=\frac{1}{\cos t}\qquad\Rightarrow\sec^{2}t=\frac{1}{\cos^{2}t}$
$\sin^{2}t+\cos^{2}t=1\qquad\Rightarrow\sin^{2}t=1-\cos^{2}t$ (not negative, anywhere)
$\displaystyle \sec^{2}t\cdot\sin^{2}t=\frac{1}{\cos^{2}t}(1-\cos^{2}t )$
$\displaystyle \sec^{2}t\cdot\sin^{2}t=\frac{1-\cos^{2}t}{\cos^{2}t}$
On the RHS, both the numerator and denominator are nonnegative,
regardless of the quadrant t belongs to,
so we can place a + sign (or leave it out).
