Question

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

• A. \( 1 \)
• B. \( 3 \)
• C. \( 4 \)
• D. \( -3 \)
• E. \( \dfrac{1}{8} \)

Hint:

When inverse trigonometric expressions are added, first use their principal value ranges. Often the extreme value of the sum immediately gives the answer.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem relies on understanding the range of the inverse cosine function. The principal value range of \(\cos^{-1}(t)\) is \([0, \pi]\).
Step 2: Key Formula or Approach:
We know that for any valid input \(t\) (where \(-1 \le t \le 1\)):
\[ 0 \le \cos^{-1}(t) \le \pi \] The given equation is the sum of three such terms. We can analyze the maximum possible value of the sum.
Step 3: Detailed Explanation:
The given equation is:
\[ \cos^{-1}x + \cos^{-1}y + \cos^{-1}z = 3\pi \] We know that the maximum value that \(\cos^{-1}(t)\) can take is \(\pi\).
Therefore, the maximum value of the left-hand side is \(\pi + \pi + \pi = 3\pi\).
The equation holds true only if each term on the left-hand side takes its maximum possible value. This means:
\[ \cos^{-1}x = \pi \] \[ \cos^{-1}y = \pi \] \[ \cos^{-1}z = \pi \] Now, we solve for x, y, and z:
If \(\cos^{-1}x = \pi\), then \(x = \cos(\pi) = -1\).
If \(\cos^{-1}y = \pi\), then \(y = \cos(\pi) = -1\).
If \(\cos^{-1}z = \pi\), then \(z = \cos(\pi) = -1\).
The question asks for the value of \(x+y+z\).
\[ x+y+z = (-1) + (-1) + (-1) = -3 \] Step 4: Final Answer:
The value of \(x+y+z\) is -3. Therefore, option (D) is the correct answer.