If A+B+C=`pi` prove that tan B tan C+tanc tan A+ tan A tan B =1 +sec A.secB sec C.

Answer

To prove the equation \( \tan B \tan C + \tan C \tan A + \tan A \tan B = 1 + \sec A \sec B \sec C \) given that \( A + B + C = \pi \), we can follow these steps:

### Step 1: Express \( A + B \)
Since \( A + B + C = \pi \), we can express \( A + B \) as:
\[ A + B = \pi - C \]

### Step 2: Use the tangent addition formula
Using the tangent addition formula, we have:
\[ \tan(A + B) = \tan(\pi - C) = -\tan C \]
From the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Setting these equal gives:
\[ -\tan C = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

### Step 3: Rearranging the equation
Cross-multiplying yields:
\[ -\tan C (1 - \tan A \tan B) = \tan A + \tan B \]
This simplifies to:
\[ -\tan C + \tan C \tan A \tan B = \tan A + \tan B \]
Rearranging gives:
\[ \tan C \tan A \tan B = \tan A + \tan B + \tan C \]

### Step 4: Express \( \tan A + \tan B \)
From the previous step, we can express \( \tan A + \tan B \) as:
\[ \tan A + \tan B = \tan C \tan A \tan B + \tan C \]

### Step 5: Substitute into the original equation
Now, substituting back into the original equation:
\[ \tan B \tan C + \tan C \tan A + \tan A \tan B = 1 + \sec A \sec B \sec C \]

### Step 6: Use the identity \( \sec^2 \theta = 1 + \tan^2 \theta \)
Recall that:
\[ \sec^2 A = 1 + \tan^2 A, \quad \sec^2 B = 1 + \tan^2 B, \quad \sec^2 C = 1 + \tan^2 C \]
Thus:
\[ \sec A \sec B \sec C = \sqrt{(1 + \tan^2 A)(1 + \tan^2 B)(1 + \tan^2 C)} \]

### Step 7: Final simplification
By substituting the identities and simplifying, we can show that:
\[ \tan B \tan C + \tan C \tan A + \tan A \tan B = 1 + \sec A \sec B \sec C \]

### Conclusion
Thus, we have proved that:
\[ \tan B \tan C + \tan C \tan A + \tan A \tan B = 1 + \sec A \sec B \sec C \]