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:

An alternate quick way to solve this is to write \(\cot^2 A\) as the reciprocal of \(\tan^2 A\):
\[ \frac{1 + \tan^2 A}{1 + \frac{1}{\tan^2 A}} = \frac{1 + \tan^2 A}{\frac{\tan^2 A + 1}{\tan^2 A}} = (1 + \tan^2 A) \cdot \frac{\tan^2 A}{1 + \tan^2 A} = \tan^2 A \] This method requires only one trigonometric identity change and is extremely elegant!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The topic of this question is Trigonometric Identities.
We are asked to simplify a given fractional algebraic expression containing tangent and cotangent terms.
We can solve this problem by applying the fundamental Pythagorean trigonometric identities or by converting all functions into their sine and cosine components.

Step 2: Key Formula or Approach:
Recall the fundamental Pythagorean identities of trigonometry:
\[ 1 + \tan^2 A = \sec^2 A \] \[ 1 + \cot^2 A = \csc^2 A \] Also recall the reciprocal definitions of these functions:
\[ \sec A = \frac{1}{\cos A} \] \[ \csc A = \frac{1}{\sin A} \] And the quotient relationship for tangent:
\[ \tan A = \frac{\sin A}{\cos A} \]

Step 3: Detailed Explanation:

• Write down the given expression:
\[ \text{Expression} = \frac{1 + \tan^2 A}{1 + \cot^2 A} \]

• Substitute the Pythagorean identities directly into the numerator and denominator:
\[ \text{Expression} = \frac{\sec^2 A}{\csc^2 A} \]

• Rewrite the secant and cosecant terms in terms of sine and cosine:
\[ \text{Expression} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} \]

• Simplify the complex fraction by multiplying the numerator by the reciprocal of the denominator:
\[ \text{Expression} = \frac{1}{\cos^2 A} \times \frac{\sin^2 A}{1} \] \[ \text{Expression} = \frac{\sin^2 A}{\cos^2 A} \]

• Convert the ratio back into the tangent function:
\[ \text{Expression} = \left(\frac{\sin A}{\cos A}\right)^2 = \tan^2 A \]

Step 4: Final Answer:
The expression \(\frac{1 + \tan^2 A}{1 + \cot^2 A}\) simplifies to \(\tan^2 A\).
Therefore, the correct option is (A).