KCET · Maths · Inverse Trigonometric Functions
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 & Solution
Correct Answer
(C) \(6\)
Step-by-step Solution
Detailed explanation
Given, \(\cos ^{-1}(x)+\cos ^{-1}(y)+\cos ^{-1}(z)=3 \pi\)
As, we know that \(0 \leq \cos ^{-1}(x) \leq \pi\)
Thus, the maximum value of \(\cos ^{-1}(x)=\pi\)
satisfies the given equation.
\(x=y=z=\cos \pi=-1\)
Now, \(x(y+z)+y(z+x)+z(x+y)\)
\(=x y+z x+y z+x y+z x+y z\)
\(=2(x y+y z+z x)\)
\(=2(1+1+1)=6\)
As, we know that \(0 \leq \cos ^{-1}(x) \leq \pi\)
Thus, the maximum value of \(\cos ^{-1}(x)=\pi\)
satisfies the given equation.
\(x=y=z=\cos \pi=-1\)
Now, \(x(y+z)+y(z+x)+z(x+y)\)
\(=x y+z x+y z+x y+z x+y z\)
\(=2(x y+y z+z x)\)
\(=2(1+1+1)=6\)
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