Understanding the Trigonometric Expression

The question asks us to simplify a given trigonometric expression involving tangent and cotangent functions. The expression is:

To simplify this, we will use fundamental trigonometric identities and algebraic manipulation. We will simplify each part of the expression separately and then combine them.

Simplifying the First Part: \( \rm \frac{1+\tan^2\theta}{1+\cot^2\theta} \)

We can use the Pythagorean identities:

• \( 1 + \tan^2\theta = \sec^2\theta \)
• \( 1 + \cot^2\theta = \csc^2\theta \)

Substitute these identities into the first part of the expression:

Now, express secant and cosecant in terms of sine and cosine:

• \( \sec\theta = \frac{1}{\cos\theta} \implies \sec^2\theta = \frac{1}{\cos^2\theta} \)
• \( \csc\theta = \frac{1}{\sin\theta} \implies \csc^2\theta = \frac{1}{\sin^2\theta} \)

Substitute these into the expression:

Dividing by a fraction is the same as multiplying by its reciprocal:

Using the identity \( \tan\theta = \frac{\sin\theta}{\cos\theta} \), we get:

So, the first part simplifies to \( \tan^2\theta \).

Simplifying the Second Part: \( \rm \left(\frac{1-\tan\theta}{1-\cot\theta}\right)^2 \)

Let's first simplify the expression inside the parenthesis: \( \rm \frac{1-\tan\theta}{1-\cot\theta} \). We can use the identity \( \cot\theta = \frac{1}{\tan\theta} \).

Substitute \( \cot\theta = \frac{1}{\tan\theta} \) into the denominator:

Combine the terms in the denominator by finding a common denominator:

Now substitute this back into the fraction:

Multiply the numerator by the reciprocal of the denominator:

Notice that \( 1-\tan\theta = -(\tan\theta-1) \). Substitute this into the expression:

Assuming \( \tan\theta \neq 1 \) (so \( \tan\theta-1 \neq 0 \)), we can cancel the term \( (\tan\theta-1) \):

Now, we need to square this result to get the second part of the original expression:

So, the second part also simplifies to \( \tan^2\theta \).

Combining the Simplified Parts

The original expression was:

We found that:

• \( \rm \frac{1+tan^2\theta}{1+cot^2\theta} = \tan^2\theta \)
• \( \rm \left(\frac{1-tan\theta}{1-cot\theta}\right)^2 = \tan^2\theta \)

Substitute these simplified forms back into the original expression:

This simplifies to:

Thus, the value of the given expression is 0.

Final Answer Derivation

By simplifying both parts of the expression using trigonometric identities and algebraic manipulation, we found that the first part is equal to \( \tan^2\theta \) and the second part is also equal to \( \tan^2\theta \). Subtracting the second part from the first part gives \( \tan^2\theta - \tan^2\theta \), which is 0.

The correct option is 0.

Revision Table: Key Trigonometric Identities

Additional Information: Restrictions

When simplifying trigonometric expressions, it is important to consider any restrictions on the angle \( \theta \). In this problem, the expressions involve \( \tan\theta \) and \( \cot\theta \).

• \( \tan\theta \) is undefined when \( \cos\theta = 0 \), i.e., \( \theta = n\pi + \frac{\pi}{2} \), where \( n \) is an integer.
• \( \cot\theta \) is undefined when \( \sin\theta = 0 \), i.e., \( \theta = n\pi \), where \( n \) is an integer.
• The term \( \frac{1-\tan\theta}{1-\cot\theta} \) required \( \tan\theta \neq 1 \). \( \tan\theta = 1 \) when \( \theta = n\pi + \frac{\pi}{4} \), where \( n \) is an integer.
• The denominator \( 1-\cot\theta \) required \( 1-\cot\theta \neq 0 \), i.e., \( \cot\theta \neq 1 \). \( \cot\theta = 1 \) when \( \theta = n\pi + \frac{\pi}{4} \), where \( n \) is an integer.

The simplification holds true for all values of \( \theta \) where the original expression is defined.