Question

Choose the correct option. Justify your choice. $\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}=$

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

Answer 1

The correct option is D

${\tan }^{2}A$

Determine the correct Option.

The explanation for the correct option:

We know that, $\cot A=\frac{1}{\tan A}$.

Therefore, ${\cot }^{2}A=\frac{1}{{\tan }^{2}A}$.

Substitute this in the given equation:

$\begin{array}{rcl}\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}& =& \frac{1+{\tan }^{2}A}{1+{\displaystyle \frac{1}{{\tan }^{2}A}}}\\ & =& \frac{1+{\tan }^{2}A}{{\displaystyle \frac{{\tan }^{2}A+1}{{\tan }^{2}A}}}\\ & =& {\tan }^{2}A\times \frac{1+{\tan }^{2}A}{1+{\tan }^{2}A}\\ & =& {\tan }^{2}A\end{array}$

Therefore, $\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}={\tan }^{2}A$.

The explanation for incorrect option:

Since the value for the given expression is $\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}={\tan }^{2}A$ is not equal to the value mentioned Option (A), Option (B) and Option (C).

Hence, Option (A), Option (B), and Option (C) are the incorrect answers.

Hence,in the correct option is (D) i.e., ${\tan }^{2}A$.