If tan A = cot B , prove that A + B = `90^(@)` .

Answer

To prove that \( A + B = 90^\circ \) given that \( \tan A = \cot B \), we can follow these steps:

### Step 1: Use the relationship between tangent and cotangent
We know that:
\[ \cot B = \frac{1}{\tan B} \]
Thus, if \( \tan A = \cot B \), we can rewrite this as:
\[ \tan A = \frac{1}{\tan B} \]

### Step 2: Rewrite the equation
From the equation \( \tan A = \frac{1}{\tan B} \), we can rearrange it to:
\[ \tan A \cdot \tan B = 1 \]

### Step 3: Use the identity for tangent
We know from trigonometric identities that:
\[ \tan(90^\circ - B) = \cot B \]
Thus, we can replace \( \cot B \) in our original equation:
\[ \tan A = \tan(90^\circ - B) \]

### Step 4: Set the angles equal
Since the tangent function is periodic, we can equate the angles:
\[ A = 90^\circ - B \]

### Step 5: Rearranging the equation
Now, we can rearrange this equation to find \( A + B \):
\[ A + B = 90^\circ \]

### Conclusion
Thus, we have proven that:
\[ A + B = 90^\circ \]