Question

If $\cos^{-1} x + \cos^{-1} y + \cos^{-1} z = 3\pi$, then $x(y + z) + y(z + x) + z(x + y)$ equals to:

• A. 0
• B. 1
• C. 6
• D. 12

Answer 1

The Correct Option is C

1. Understand the problem:

Given cos⁻¹x + cos⁻¹y + cos⁻¹z = 3π, we need to find the value of x(y+z) + y(z+x) + z(x+y).

2. Analyze the given equation:

The maximum value of cos⁻¹θ for any θ ∈ [-1,1] is π. The sum equals 3π only if each term equals π.

Thus, cos⁻¹x = cos⁻¹y = cos⁻¹z = π ⇒ x = y = z = -1.

3. Substitute x = y = z = -1:

The expression becomes:

(-1)(-1-1) + (-1)(-1-1) + (-1)(-1-1) = (-1)(-2) + (-1)(-2) + (-1)(-2) = 2 + 2 + 2 = 6

Correct Answer: (C) 6

Answer 2

Given that \( \cos^{-1}x + \cos^{-1}y + \cos^{-1}z = 3\pi \).

The range of \( \cos^{-1}(\theta) \) is \( [0, \pi] \). 

Since the sum of three angles in the range \( [0, \pi] \) is \( 3\pi \), each angle must be exactly \( \pi \).

Therefore:

\[ \cos^{-1}x = \pi, \quad \cos^{-1}y = \pi, \quad \cos^{-1}z = \pi \]

This implies:

\[ x = \cos(\pi) = -1, \quad y = \cos(\pi) = -1, \quad z = \cos(\pi) = -1 \]

Now let's substitute these values into the expression \( x(y+z) + y(z+x) + z(x+y) \):

\[ x(y+z) + y(z+x) + z(x+y) = (-1)((-1) + (-1)) + (-1)((-1) + (-1)) + (-1)((-1) + (-1)) \] \[ = (-1)(-2) + (-1)(-2) + (-1)(-2) \] \[ = 2 + 2 + 2 \] \[ = 6 \]

Therefore, \( x(y+z) + y(z+x) + z(x+y) = 6 \).