Question:
$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}$ is equal to
(a) $\sec ^{2} A$
(b) $-1$
(c) $\cot ^{2} A$
(d) $\tan ^{2} A$
Solution:
Given:
$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}$
$=\frac{1+\frac{\sin ^{2} A}{\cos ^{2} A}}{1+\frac{\cos ^{2} A}{\sin ^{2} A}}$
$=\frac{\frac{\cos ^{2} A+\sin ^{2} A}{\cos ^{2} A}}{\frac{\sin ^{2} A+\cos ^{2} A}{\sin ^{2} A}}$
$=\frac{\frac{1}{\cos ^{2} A}}{\frac{1}{\sin ^{2} A}}$
$=\frac{\sin ^{2} A}{\cos ^{2} A}$
$=\tan ^{2} A$
Therefore, the correct option is (d).
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