If ` tan(pi cos theta )= cot (pi sin theta ) ` ,then ` cos^(2)(theta -pi//4) ` is equal to

Answer

To solve the equation \( \tan(\pi \cos \theta) = \cot(\pi \sin \theta) \) and find the value of \( \cos^2\left(\theta - \frac{\pi}{4}\right) \), we can follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ \tan(\pi \cos \theta) = \cot(\pi \sin \theta) \]
We know that \( \cot x = \frac{1}{\tan x} \), so we can rewrite the equation as:
\[ \tan(\pi \cos \theta) = \frac{1}{\tan(\pi \sin \theta)} \]
This implies:
\[ \tan(\pi \cos \theta) \tan(\pi \sin \theta) = 1 \]

### Step 2: Use the Identity
Using the identity \( \tan A \tan B = 1 \) leads us to:
\[ \pi \cos \theta + \pi \sin \theta = \frac{\pi}{2} + n\pi \quad \text{for some integer } n \]
This can be simplified to:
\[ \cos \theta + \sin \theta = \frac{1}{2} + \frac{n}{\pi} \]

### Step 3: Solve for \( \cos \theta + \sin \theta \)
We can express \( \cos \theta + \sin \theta \) in terms of a single cosine function:
\[ \cos \theta + \sin \theta = \sqrt{2} \cos\left(\theta - \frac{\pi}{4}\right) \]
Thus, we have:
\[ \sqrt{2} \cos\left(\theta - \frac{\pi}{4}\right) = \frac{1}{2} + \frac{n}{\pi} \]

### Step 4: Isolate \( \cos\left(\theta - \frac{\pi}{4}\right) \)
Dividing both sides by \( \sqrt{2} \):
\[ \cos\left(\theta - \frac{\pi}{4}\right) = \frac{1}{2\sqrt{2}} + \frac{n}{\sqrt{2}\pi} \]

### Step 5: Find \( \cos^2\left(\theta - \frac{\pi}{4}\right) \)
Now, we need to find \( \cos^2\left(\theta - \frac{\pi}{4}\right) \):
\[ \cos^2\left(\theta - \frac{\pi}{4}\right) = \left(\frac{1}{2\sqrt{2}} + \frac{n}{\sqrt{2}\pi}\right)^2 \]
Calculating this gives:
\[ \cos^2\left(\theta - \frac{\pi}{4}\right) = \left(\frac{1}{2\sqrt{2}}\right)^2 = \frac{1}{8} \]

### Final Answer
Thus, the value of \( \cos^2\left(\theta - \frac{\pi}{4}\right) \) is:
\[ \frac{1}{8} \]