Question

\(\frac{1 + \tan^2 \theta}{1 + \cot^2 \theta}\) equals ?

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

Hint:

1. Use Pythagorean identities: \(1 + \tan^2 \theta = \sec^2 \theta\) \(1 + \cot^2 \theta = \csc^2 \theta\) 2. The expression becomes \(\frac{\sec^2 \theta}{\csc^2 \theta}\). 3. Convert to sines and cosines: \(\sec^2 \theta = 1/\cos^2 \theta\) \(\csc^2 \theta = 1/\sin^2 \theta\) 4. So, \(\frac{1/\cos^2 \theta}{1/\sin^2 \theta} = \frac{\sin^2 \theta}{\cos^2 \theta}\). 5. Recognize that \(\frac{\sin^2 \theta}{\cos^2 \theta} = (\frac{\sin \theta}{\cos \theta})^2 = \tan^2 \theta\).

Answer 1

The Correct Option is D

Concept: This problem uses fundamental Pythagorean trigonometric identities: (A) \(1 + \tan^2 \theta = \sec^2 \theta\) (B) \(1 + \cot^2 \theta = \csc^2 \theta\) (or \(\text{cosec}^2 \theta\)) And also the reciprocal identities: \(\sec \theta = \frac{1}{\cos \theta}\), \(\csc \theta = \frac{1}{\sin \theta}\), and the quotient identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). Step 1: Apply Pythagorean identities to the numerator and denominator Numerator: \(1 + \tan^2 \theta = \sec^2 \theta\) Denominator: \(1 + \cot^2 \theta = \csc^2 \theta\) Substitute these into the expression: \[ \frac{1 + \tan^2 \theta}{1 + \cot^2 \theta} = \frac{\sec^2 \theta}{\csc^2 \theta} \] Step 2: Express \(\sec^2 \theta\) and \(\csc^2 \theta\) in terms of \(\sin \theta\) and \(\cos \theta\) We know:
\(\sec \theta = \frac{1}{\cos \theta}\), so \(\sec^2 \theta = \frac{1}{\cos^2 \theta}\)
\(\csc \theta = \frac{1}{\sin \theta}\), so \(\csc^2 \theta = \frac{1}{\sin^2 \theta}\) Substitute these into the expression from Step 1: \[ \frac{\sec^2 \theta}{\csc^2 \theta} = \frac{\frac{1}{\cos^2 \theta}}{\frac{1}{\sin^2 \theta}} \] Step 3: Simplify the complex fraction To divide fractions, multiply by the reciprocal of the denominator: \[ \frac{\frac{1}{\cos^2 \theta}}{\frac{1}{\sin^2 \theta}} = \frac{1}{\cos^2 \theta} \times \frac{\sin^2 \theta}{1} = \frac{\sin^2 \theta}{\cos^2 \theta} \] Step 4: Use the quotient identity for tangent We know that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). Therefore, \(\tan^2 \theta = \left(\frac{\sin \theta}{\cos \theta}\right)^2 = \frac{\sin^2 \theta}{\cos^2 \theta}\). So, the expression simplifies to: \[ \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta \] The value of the expression is \(\tan^2 \theta\). This matches option (4).