A3. Which expression is equivalent to the expression $\frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{-1} x+\cos ^{-1} x} \cdot \frac{\tan x}{\cos x-\sin x}$ ?
(A) $-\cos x$
(B) $-\sin x$
(C) $-\sin ^{2} x$
(D) $-\cos ^{2} x$
(E) $\cos x$
A3. The numerator of the first fraction can be factored as the difference of squares $\sin ^{2} x-\cos ^{2} x=(\sin x-\cos x)(\sin x+$ $\cos x)$, and the denominator can be written as the sum of reciprocals $\sin ^{-1} x+\cos ^{-1} x=\frac{1}{\sin x}+\frac{1}{\cos x}$, and the resulting fractions are added. The numerator of the second fraction is written as $\tan x=\frac{\sin x}{\cos x}$, and in the denominator of the second fraction, we factor out $\cos x-\sin x=-(\sin x-\cos x)$. This leads to the simplification of the fractions and we get the result $-\sin ^{2} x$. The correct answer is C.