If `x=2tanalpha,y=2cotalpha`, then `16((1)/(4+x^(2))+(1)/(4+y^(2)))=?`

Options

A. 1
B. 2
C. 4
D. 8

Answer

To solve the problem step by step, we start with the given expressions for \( x \) and \( y \):

1. **Given Values**:
\[ x = 2 \tan \alpha, \quad y = 2 \cot \alpha \]

2. **Substituting into the Expression**:
We need to evaluate:
\[ 16 \left( \frac{1}{4 + x^2} + \frac{1}{4 + y^2} \right) \]
First, we substitute \( x \) and \( y \):
\[ x^2 = (2 \tan \alpha)^2 = 4 \tan^2 \alpha, \quad y^2 = (2 \cot \alpha)^2 = 4 \cot^2 \alpha \]

3. **Rewriting the Expression**:
Substitute \( x^2 \) and \( y^2 \) into the expression:
\[ 16 \left( \frac{1}{4 + 4 \tan^2 \alpha} + \frac{1}{4 + 4 \cot^2 \alpha} \right) \]

4. **Factoring Out the Common Terms**:
Factor out the 4 from the denominators:
\[ = 16 \left( \frac{1}{4(1 + \tan^2 \alpha)} + \frac{1}{4(1 + \cot^2 \alpha)} \right) \]
This simplifies to:
\[ = 16 \cdot \frac{1}{4} \left( \frac{1}{1 + \tan^2 \alpha} + \frac{1}{1 + \cot^2 \alpha} \right) \]
\[ = 4 \left( \frac{1}{1 + \tan^2 \alpha} + \frac{1}{1 + \cot^2 \alpha} \right) \]

5. **Using Trigonometric Identities**:
We know from trigonometric identities that:
\[ 1 + \tan^2 \alpha = \sec^2 \alpha, \quad 1 + \cot^2 \alpha = \csc^2 \alpha \]
Therefore, we can rewrite the expression as:
\[ = 4 \left( \frac{1}{\sec^2 \alpha} + \frac{1}{\csc^2 \alpha} \right) \]

6. **Simplifying Further**:
Using the identities \( \frac{1}{\sec^2 \alpha} = \cos^2 \alpha \) and \( \frac{1}{\csc^2 \alpha} = \sin^2 \alpha \):
\[ = 4 \left( \cos^2 \alpha + \sin^2 \alpha \right) \]

7. **Final Result**:
We know from the Pythagorean identity that:
\[ \cos^2 \alpha + \sin^2 \alpha = 1 \]
Thus, we have:
\[ = 4 \cdot 1 = 4 \]

Therefore, the final answer is:
\[ \boxed{4} \]