Question

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

• A. ${\sec }^{2}A$
• B. $-1$
• C. ${\cot }^{2}A$
• D. ${\tan }^{2}A$

Answer 1

The correct option is D ${\tan }^{2}A$

$\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}$
$=\frac{{\sec }^{2}A}{cose{c}^{2}A}[\because 1+ta{n}^2\theta =se{c}^2\theta and1+co{t}^2\theta =cose{c}^2\theta ]$
$=\frac{1}{{\cos }^{2}A}\times \frac{{\sin }^{2}A}{1}$
$={\tan }^{2}A$