Question

The value of $\dfrac{1+\cot^2 \theta}{1+\tan^2 \theta}$ will be:
 

• A. $\sec^2 \theta$
• B. $\csc^2 \theta$
• C. $\tan^2 \theta$
• D. $\cot^2 \theta$

Hint:

Whenever you see expressions like $1+\cot^2 \theta$ or $1+\tan^2 \theta$, immediately recall the Pythagorean identities: $1+\cot^2 \theta = \csc^2 \theta$ and $1+\tan^2 \theta = \sec^2 \theta$.

Answer 1

The Correct Option is B

Step 1: Recall Pythagorean identities
\[ 1 + \cot^2 \theta = \csc^2 \theta \text{and} 1 + \tan^2 \theta = \sec^2 \theta \]

Step 2: Substitute these identities into the given expression
\[ \frac{1+\cot^2 \theta}{1+\tan^2 \theta} = \frac{\csc^2 \theta}{\sec^2 \theta} \]

Step 3: Simplify the ratio
\[ \frac{\csc^2 \theta}{\sec^2 \theta} = \frac{\dfrac{1}{\sin^2 \theta}}{\dfrac{1}{\cos^2 \theta}} = \frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta \] Wait, let's carefully check: - Numerator: $1+\cot^2 \theta = \csc^2 \theta$ - Denominator: $1+\tan^2 \theta = \sec^2 \theta$ So: \[ \frac{\csc^2 \theta}{\sec^2 \theta} = \frac{\dfrac{1}{\sin^2 \theta}}{\dfrac{1}{\cos^2 \theta}} = \frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta \] Step 4: Final Answer
\[ \boxed{\dfrac{1+\cot^2 \theta}{1+\tan^2 \theta} = \cot^2 \theta} \] Thus, the correct option is (D).