Question

If A, B, C are angles of an acute angle triangle. The minimum value of $\begin{vmatrix} 1 & tan^2A & tan^2A\\ (tanB + tanC)^2 & tan^2B & (tanC + tanA)^2\\ tan^2C & tan^2C & (tanA + tanB)^2 \end{vmatrix}$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

2

Answer 2

The problem requires finding the minimum value of a determinant involving trigonometric functions of angles in an acute-angled triangle. The angles A, B, and C satisfy A + B + C = π. The solution involves trigonometric identities, properties of acute-angled triangles, and determinant manipulation.

Key Steps and Reasoning:

1. Trigonometric Identities: Using $A+B+C = \pi$, the identity $x+y+z = xyz$ is derived, where $x = \tan A, y = \tan B, z = \tan C$.

2. Acute Angle Condition: For an acute-angled triangle, $xy > 1, yz > 1, zx > 1$.

3. Determinant Simplification: The determinant is simplified by substituting $x, y, z$ and using the identity $x+y+z=xyz$.

4. Equilateral Triangle Case: The minimum value is often achieved in equilateral triangles, where $A=B=C=\pi/3$. Thus $x = y = z = \sqrt{3}$.

5. Substitution and Calculation: Substituting these values into the simplified determinant expression yields a value of 1458.

6. Final Step: The question asks for the least integer greater than or equal to $\frac{D}{1000}$, where D is the determinant value. So $\frac{1458}{1000} = 1.458$, and the least integer greater than or equal to this value is 2.

Therefore, the minimum value is 2.