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: Recall the range of \( \cos^{-1}t \).
For any real number \( t \in [-1,1] \), \[ \cos^{-1}t \in [0,\pi] \] So each of the three angles \[ \cos^{-1}x,\quad \cos^{-1}y,\quad \cos^{-1}z \] lies between \( 0 \) and \( \pi \).

Step 2: Use the maximum possible sum.
Since each inverse cosine value is at most \( \pi \), the maximum possible value of \[ \cos^{-1}x+\cos^{-1}y+\cos^{-1}z \] is \[ \pi+\pi+\pi=3\pi \] But the question states that the sum is exactly \[ 3\pi \]

Step 3: Conclude that each term must be maximum.
A sum of three numbers can equal its maximum possible value only when each number individually equals its maximum value.
Therefore, \[ \cos^{-1}x=\pi,\qquad \cos^{-1}y=\pi,\qquad \cos^{-1}z=\pi \]

Step 4: Convert back to \( x,y,z \).
Now use the fact that \[ \cos^{-1}t=\pi \quad \Longrightarrow \quad t=\cos\pi=-1 \] Hence, \[ x=-1,\qquad y=-1,\qquad z=-1 \]

Step 5: Find the required sum.
So, \[ x+y+z=-1-1-1=-3 \]

Step 6: Verify quickly.
Substituting back: \[ \cos^{-1}(-1)+\cos^{-1}(-1)+\cos^{-1}(-1)=\pi+\pi+\pi=3\pi \] So the condition is satisfied exactly.

Step 7: Final conclusion.
Therefore, \[ \boxed{x+y+z=-3} \] Hence, the correct option is \[ \boxed{(4)\ -3} \]