Question

\(\frac{1 + \tan^2 A}{1 + \cot^2 A}\) equals to:

• A. \(\tan^2 A\)
• B. \(-1\)
• C. \(-\tan^2 A\)
• D. \(\cot^2 A\)

Hint:

A quick shortcut for such fractions is to remember that \(\cot A = 1/\tan A\). Thus, \(1 + \cot^2 A = 1 + \frac{1}{\tan^2 A} = \frac{\tan^2 A + 1}{\tan^2 A}\). Dividing the numerator by this immediately gives \(\tan^2 A\).

Answer 1

The Correct Option is A

To solve the expression \(\frac{1 + \tan^2 A}{1 + \cot^2 A}\), we need to use trigonometric identities. First, we recall the Pythagorean identity for tangent and cotangent:

• \(\tan^2 A = \sec^2 A - 1\)
• \(\cot^2 A = \csc^2 A - 1\)

Let's simplify the expression step-by-step.

1. We know: \(1 + \tan^2 A = \sec^2 A\)
2. Similarly, \(1 + \cot^2 A = \csc^2 A\)
3. Substitute these into the expression: \(\frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\csc^2 A}\)
4. Rewrite \(\sec^2 A\) and \(\csc^2 A\) in terms of sine and cosine:
   • \(\sec^2 A = \frac{1}{\cos^2 A}\)
   • \(\csc^2 A = \frac{1}{\sin^2 A}\)
5. Substitute back into the expression: \(\frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{\sin^2 A}{\cos^2 A}\)
6. Simplify the fraction: \(\frac{\sin^2 A}{\cos^2 A} = \tan^2 A\)

Therefore, the expression \(\frac{1 + \tan^2 A}{1 + \cot^2 A}\) simplifies to \(\tan^2 A\). Hence, the correct answer is the option:

\(\tan^2 A\)