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 ${\tan }^{2}A$

We know that $\tan \theta .\cot \theta =1$ $...........\left(1\right)$

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

$=\frac{1+{\tan }^{2}A}{1+{\left(\frac{1}{\tan A}\right)}^2}$ $\left(From\right(1\left)\right)$

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

$=\left(\frac{1+{\tan }^{2}A}{{\tan }^{2}A+1}\right){\tan }^{2}A$

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

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