Prove that `cot(pi/4-2cot^(- 1)3)=7`

Explanation

To prove that \( \cot\left(\frac{\pi}{4} - 2\cot^{-1}(3)\right) = 7 \), we will start with the left-hand side (LHS) and simplify it step by step.

### Step 1: Rewrite the expression
We know that:
\[ \cot\left(\frac{\pi}{4} - 2\cot^{-1}(3)\right) \]
We can use the identity for \( \cot(A - B) \):
\[ \cot(A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A} \]
Here, \( A = \frac{\pi}{4} \) and \( B = 2\cot^{-1}(3) \).

### Step 2: Calculate \( \cot\left(\frac{\pi}{4}\right) \)
We know that:
\[ \cot\left(\frac{\pi}{4}\right) = 1 \]

### Step 3: Find \( \cot(2\cot^{-1}(3)) \)
Using the double angle formula for cotangent:
\[ \cot(2\theta) = \frac{\cot^2(\theta) - 1}{2\cot(\theta)} \]
Let \( \theta = \cot^{-1}(3) \). Then \( \cot(\theta) = 3 \).

Substituting into the formula:
\[ \cot(2\cot^{-1}(3)) = \frac{3^2 - 1}{2 \cdot 3} = \frac{9 - 1}{6} = \frac{8}{6} = \frac{4}{3} \]

### Step 4: Substitute back into the cotangent difference formula
Now we can substitute back into our expression:
\[ \cot\left(\frac{\pi}{4} - 2\cot^{-1}(3)\right) = \frac{\cot\left(\frac{\pi}{4}\right) \cot(2\cot^{-1}(3)) + 1}{\cot(2\cot^{-1}(3)) - \cot\left(\frac{\pi}{4}\right)} \]
Substituting the values we found:
\[ = \frac{1 \cdot \frac{4}{3} + 1}{\frac{4}{3} - 1} \]

### Step 5: Simplify the numerator and denominator
Calculating the numerator:
\[ 1 \cdot \frac{4}{3} + 1 = \frac{4}{3} + \frac{3}{3} = \frac{7}{3} \]

Calculating the denominator:
\[ \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3} \]

### Step 6: Final calculation
Now substituting these back:
\[ \cot\left(\frac{\pi}{4} - 2\cot^{-1}(3)\right) = \frac{\frac{7}{3}}{\frac{1}{3}} = 7 \]

Thus, we have shown that:
\[ \cot\left(\frac{\pi}{4} - 2\cot^{-1}(3)\right) = 7 \]

### Conclusion
Hence, we have proved that:
\[ \cot\left(\frac{\pi}{4} - 2\cot^{-1}(3)\right) = 7 \]

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