Question

$\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}=$ _____

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

Answer 1

Answer: (D)

${\tan }^{2}A$

${\sin }^{2}A+{\cos }^{2}A=1$

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

$=\frac{\frac{1}{{\cos }^{2}A}}{\frac{1}{{\sin }^{2}A}}=\frac{{\sin }^{2}A}{{\cos }^{2}A}={\tan }^{2}A$
So, ${\tan }^{2}A$ is correct.