1+tan2 A1+cot2 Ais equal to

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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