Q80 / 180 · KCET 2024 · Mathematics
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$1$
- Correct answerC$6$
- D$12$
Solution
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, $x(y+z)+y(z+x)+z(x+y)$
$=x y+z x+y z+x y+z x+y z$
$=2(x y+y z+z x)$
$=2(1+1+1)=6$
Hence, the correct answer is option: 'C'.