Question

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

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

Answer 1

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

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

= $\frac{1+\frac{si{n}^{2}A}{co{s}^{2}A}}{1+\frac{co{s}^{2}A}{si{n}^{2}A}}$

= $\frac{\frac{co{s}^{2}A+si{n}^{2}A}{co{s}^{2}A}}{\frac{si{n}^{2}A+co{s}^{2}A}{si{n}^{2}A}}$

= $\frac{{\sin }^{2}A}{{\cos }^{2}A}$

= ${\tan }^{2}A$