Question

The value of $\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}$ is equal to

• A. ${\tan }^{2}A$
• B. ${\sec }^{2}A$
• C. $\cos A$
• D. $\sin A$

Answer 1

Answer: (A)

${\tan }^{2}A$

it can be written as

$\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}=\frac{{\sec }^{2}A}{{\text{cosec}}^{2}A}$
$=\frac{\frac{1}{{\cos }^{2}A}}{\frac{1}{{\sin }^{2}A}}=\frac{{\sin }^{2}A}{{\cos }^{2}A}={\tan }^{2}A$

Therefore, Answer is ${\tan }^{2}A$