If A, B and C are angles of a triangle such that `angleA` is obtuse, then show tan B `tan C lt 1.`

Answer

To show that \( \tan B \tan C < 1 \) given that \( A \) is an obtuse angle in triangle \( ABC \), we can follow these steps:

### Step 1: Understand the properties of angles in a triangle
In any triangle, the sum of the angles is \( \pi \) radians (or \( 180^\circ \)). Therefore, we have:
\[ A + B + C = \pi \]

### Step 2: Express angle A in terms of B and C
Since \( A \) is obtuse, we know:
\[ \frac{\pi}{2} < A < \pi \]
This implies:
\[ A = \pi - (B + C) \]

### Step 3: Set up the inequality based on the obtuse angle
From the inequality \( A > \frac{\pi}{2} \), we can write:
\[ \pi - (B + C) > \frac{\pi}{2} \]
Rearranging this gives:
\[ B + C < \frac{\pi}{2} \]

### Step 4: Apply the tangent function
Now, we will apply the tangent function to both sides of the inequality:
\[ \tan(B + C) < \tan\left(\frac{\pi}{2}\right) \]
Since \( \tan\left(\frac{\pi}{2}\right) \) is undefined, we need to analyze \( \tan(B + C) \) further.

Using the tangent addition formula:
\[ \tan(B + C) = \frac{\tan B + \tan C}{1 - \tan B \tan C} \]
We know that \( B + C < \frac{\pi}{2} \) implies \( \tan(B + C) < \infty \). Therefore, we can conclude:
\[ 1 - \tan B \tan C > 0 \]
This leads us to:
\[ \tan B \tan C < 1 \]

### Conclusion
Thus, we have shown that if \( A \) is an obtuse angle in triangle \( ABC \), then:
\[ \tan B \tan C < 1 \]

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