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:

You can also solve this by converting \(\cot^2 A\) into its reciprocal form, \(\frac{1}{\tan^2 A}\):
\[ \frac{1 + \tan^2 A}{1 + \frac{1}{\tan^2 A}} = \frac{1 + \tan^2 A}{\frac{\tan^2 A + 1}{\tan^2 A}} = (1 + \tan^2 A) \times \frac{\tan^2 A}{1 + \tan^2 A} = \tan^2 A \] This method requires fewer identity changes and solves the expression in just two steps!

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 expression containing trigonometric terms.
The expression can be simplified by substituting the fundamental Pythagorean identities or by converting all terms into sine and cosine functions.

Step 2: Key Formula or Approach: Recall the basic Pythagorean identities:
\[ 1 + \tan^2 A = \sec^2 A \] \[ 1 + \cot^2 A = \csc^2 A \] Also, recall the relationships between the reciprocal trigonometric functions:
\[ \sec A = \frac{1}{\cos A} \] \[ \csc A = \frac{1}{\sin A} \] And the definition of the tangent function:
\[ \tan A = \frac{\sin A}{\cos A} \]

Step 3: Detailed Explanation:

• Substitute the Pythagorean identities directly into the given expression:
\[ \frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\csc^2 A} \]

• Express \(\sec^2 A\) and \(\csc^2 A\) in terms of sine and cosine:
\[ \frac{\sec^2 A}{\csc^2 A} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} \]

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

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

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