Question

$\frac{1+ta{n}^{2}A}{1+co{t}^{2}A}$ is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$\frac{1+ta{n}^{2}A}{1+co{t}^{2}A}$
$=\frac{se{c}^{2}A}{cose{c}^{2}A}$
$=ta{n}^{2}A$