Question

 If ${{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=3\pi $ then the value of xy +yz +zx is equal to:
A. 0
B. 1
C. 3
D. -3

Answer 1

Hint: In the above expression ${{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=3\pi $which will be possible only when $cos^-1x$ = π, $cos^-1y$ = π and $cos^-1z$ = π. If $cos^-1x$ = π then x is equal to -1. Similarly y and z is also equal to -1 and substitute these x, y and z values in xy +yz +zx.

Complete step by step answer:
The range of ${{\cos }^{-1}}x$is from 0 to π or$0\le {{\cos }^{-1}}x\le \pi $.
The expression given in question is:
${{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=3\pi $
As we have shown above that the range of ${{\cos }^{-1}}x$is from 0 to π or $0\le {{\cos }^{-1}}x\le \pi $ so the minimum value of ${{\cos }^{-1}}x$ is 0 and maximum value of ${{\cos }^{-1}}x$ is π. So, in the above given equation is possible when individually${{\cos }^{-1}}x,{{\cos }^{-1}}y,{{\cos }^{-1}}z$ is equal to π.
From the inverse trigonometric functions, we know that, ${{\cos }^{-1}}x=\pi $ then x = -1. Similarly, y and z are also equal to -1.
So, the values of x = y = z = -1.
Substituting the values of x, y and z in the expression xy +yz +zx we get,
$\begin{align}
  & \left( -1 \right)\left( -1 \right)+\left( -1 \right)\left( -1 \right)+\left( -1 \right)\left( -1 \right) \\
 & =-3 \\
\end{align}$
Hence, the value of the expression xy +yz +zx is 3.
Hence, the correct option is (c).

Note: You might think of doing the above problem as:
${{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=3\pi $
Subtracting ${{\cos }^{-1}}z$ both the sides and then taking cosine both the sides we get,
$\cos \left( {{\cos }^{-1}}x+{{\cos }^{-1}}y \right)=\cos \left( 3\pi -{{\cos }^{-1}}z \right)$
Using identity $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ in the above equation we get,
$\begin{align}
  & \cos \left( {{\cos }^{-1}}x \right)\cos \left( {{\cos }^{-1}}y \right)-\sin \left( {{\cos }^{-1}}x \right)\sin \left( {{\cos }^{-1}}y \right)=-z \\
 & \Rightarrow xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}}=-z \\
 & \Rightarrow xy+z=\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \\
 & \\
\end{align}$
Squaring both the sides we get,
$\begin{align}
  & {{\left( xy+z \right)}^{2}}=\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right) \\
 & \Rightarrow {{x}^{2}}{{y}^{2}}+2xyz+{{z}^{2}}=1-\left( {{x}^{2}}+{{y}^{2}} \right)+{{x}^{2}}{{y}^{2}} \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1 \\
\end{align}$
Now, resolving the above expression in terms of xy + yz + zx is pretty difficult. But the question could also ask you to evaluate the given expression ${{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=3\pi $ and write it in the form of${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1$.