Answer
$\displaystyle \tan^{2}t=\frac{\sin^{2}t}{1-\sin^{2}t}$
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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$\displaystyle \tan t=\frac{\sin t}{\cos t}\qquad\Rightarrow\tan^{2}t=\frac{\sin^{2}t}{\cos^{2}t}\quad(*)$
$\sin^{2}t+\cos^{2}t=1\qquad\Rightarrow\cos^{2}t=1-\sin^{2}t $(never negative)
Substitute $\cos^{2}t$ in (*)
$\displaystyle \tan^{2}t=\frac{\sin^{2}t}{\cos^{2}t}$
$\displaystyle \tan^{2}t=\frac{\sin^{2}t}{1-\sin^{2}t}$
We don't need the sign table here, as
the LHS is a square of a number (not negative)
the RHS is a ratio of nonnegatives (not negative)
so, if defined, regardless of quadrant,
we may put a + sign (or leave it out).
