If `(cot alpha+cot(270^(@)+alpha))/(cot alpha-cot(270^(@)+alpha))-2 cos(135^(@)+alpha)cos(315^(@)-alpha)=lambda`, where `alpha in (0, (pi)/(2))`, then `lambda=`

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the equation

\[ \frac{\cot \alpha + \cot(270^\circ + \alpha)}{\cot \alpha - \cot(270^\circ + \alpha)} - 2 \cos(135^\circ + \alpha) \cos(315^\circ - \alpha) = \lambda, \]

where \(\alpha \in (0, \frac{\pi}{2})\), we will simplify the left-hand side step by step.

### Step 1: Simplify \(\cot(270^\circ + \alpha)\)

Using the cotangent identity, we know that:

\[ \cot(270^\circ + \alpha) = \tan(\alpha). \]

Thus, we can rewrite the expression as:

\[ \frac{\cot \alpha + \tan \alpha}{\cot \alpha - \tan \alpha}. \]

### Step 2: Rewrite \(\tan\) in terms of \(\cot\)

Recall that \(\tan \alpha = \frac{1}{\cot \alpha}\). Therefore, we can substitute this into our expression:

\[ \frac{\cot \alpha + \frac{1}{\cot \alpha}}{\cot \alpha - \frac{1}{\cot \alpha}}. \]

### Step 3: Simplify the fractions

Now, we can simplify the numerator and denominator:

- **Numerator**:

\[ \cot \alpha + \frac{1}{\cot \alpha} = \frac{\cot^2 \alpha + 1}{\cot \alpha} = \frac{\csc^2 \alpha}{\cot \alpha}. \]

- **Denominator**:

\[ \cot \alpha - \frac{1}{\cot \alpha} = \frac{\cot^2 \alpha - 1}{\cot \alpha} = \frac{\cos^2 \alpha - \sin^2 \alpha}{\sin^2 \alpha \cos^2 \alpha}. \]

Thus, the entire expression becomes:

\[ \frac{\csc^2 \alpha}{\cot^2 \alpha - 1} = \frac{1}{\sin^2 \alpha} \cdot \frac{\sin^2 \alpha \cos^2 \alpha}{\cos^2 \alpha - \sin^2 \alpha}. \]

### Step 4: Simplify the cosine terms

Next, we need to simplify \( -2 \cos(135^\circ + \alpha) \cos(315^\circ - \alpha) \):

Using the cosine addition formula:

\[ \cos(135^\circ + \alpha) = -\frac{1}{\sqrt{2}} \cos \alpha - \frac{1}{\sqrt{2}} \sin \alpha, \]

and

\[ \cos(315^\circ - \alpha) = \frac{1}{\sqrt{2}} \cos \alpha + \frac{1}{\sqrt{2}} \sin \alpha. \]

Now, multiplying these two:

\[ -2 \left(-\frac{1}{\sqrt{2}} \cos \alpha - \frac{1}{\sqrt{2}} \sin \alpha\right) \left(\frac{1}{\sqrt{2}} \cos \alpha + \frac{1}{\sqrt{2}} \sin \alpha\right). \]

This expands to:

\[ -2 \left(-\frac{1}{2}(\cos^2 \alpha - \sin^2 \alpha)\right) = \cos^2 \alpha - \sin^2 \alpha. \]

### Step 5: Combine the results

Now, we can combine the results from steps 3 and 4:

\[ \frac{1}{\cos^2 \alpha - \sin^2 \alpha} - (\cos^2 \alpha - \sin^2 \alpha) = \lambda. \]

### Step 6: Final simplification

This leads to:

\[ \frac{1 - (\cos^2 \alpha - \sin^2 \alpha)^2}{\cos^2 \alpha - \sin^2 \alpha} = \lambda. \]

### Step 7: Conclusion

We can analyze the limits of \(\lambda\) as \(\alpha\) varies from \(0\) to \(\frac{\pi}{2}\). The value of \(\lambda\) will be positive and will vary from \(2\) to \(\infty\).

Thus, the final answer is:

\[ \lambda \in (2, \infty). \]