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MHT CET · Maths · Inverse Trigonometric Functions

If \(\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi\), then the value of \(x^2+y^2+z^2-2 x y z\) is

  1. A 3
  2. B 2
  3. C 1
  4. D 5
Verified Solution

Answer & Solution

Correct Answer

(D) 5

Step-by-step Solution

Detailed explanation

Since, \(0 \leq \cos ^{-1} x \leq \pi\)
\(\therefore \quad \cos ^{-1} x\) cannot be greater than \(\pi\)
\(\therefore \quad \cos ^{-1} x=\cos ^{-1} y=\cos ^{-1} x=\pi\)
Therefore, \(x=y=\mathrm{z}=-1\)
Putting these values in the given expression, we get
\(\begin{aligned}
& x^2+y^2+z^2-2 x y z \\
& =(-1)^2+(-1)^2+(-1)^2-2(-1)(-1)(-1) \\
& =1+1+1-2(-1) \\
& =3+2 \\
& =5
\end{aligned}\)