Question

If ${{\sin }^{-1}}x+{{\sin }^{-1}}y+{{\sin }^{-1}}z=\dfrac{\pi }{2}$, then the value of ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz$ is
A. 0
B. 1
C. 2
D. 3

Answer 1

Hint: We first take the inverse forms in simple angle form. Then we take ratio cos and expand the series. The corresponding ratio gives the algebraic relation between $x,y,z$. We take the square and find the value of ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz$.

Complete step by step answer:
The given equation forms ${{\sin }^{-1}}x+{{\sin }^{-1}}y=\dfrac{\pi }{2}-{{\sin }^{-1}}z$.
We assume ${{\sin }^{-1}}x=a;{{\sin }^{-1}}y=b;{{\sin }^{-1}}z=c$. Therefore, $a+b=\dfrac{\pi }{2}-c$.We can also write
${{\sin }^{-1}}m={{\cos }^{-1}}\sqrt{1-{{m}^{2}}}\\
\Rightarrow {{\sin }^{-1}}n={{\cos }^{-1}}\sqrt{1-{{n}^{2}}}$.
We take the trigonometric ratio of cos for the relation $a+b=\dfrac{\pi }{2}-c$.
$\cos \left( a+b \right)=\cos \left( \dfrac{\pi }{2}-c \right)=\sin c$.
From ${{\sin }^{-1}}x={{\cos }^{-1}}\sqrt{1-{{x}^{2}}}=a;{{\sin }^{-1}}y={{\cos }^{-1}}\sqrt{1-{{y}^{2}}}=b;{{\sin }^{-1}}z=c$, we get
\[\sin a=x,\cos \left( a \right)=\sqrt{1-{{x}^{2}}};\sin \left( b \right)=\sqrt{1-{{y}^{2}}};\sin \left( c \right)=z\].

We now use the formulas of $\cos \left( m+n \right)=\cos m\cos n-\sin m\sin n$.Therefore, we get
\[\cos \left( a+b \right)=\sin c \\
\Rightarrow \cos a\cos b-\sin a\sin b=\sin c \\
\Rightarrow \sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}}-xy=z \\
\Rightarrow z+xy=\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \\ \]
We now take square both sides to get
\[z+xy=\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \\
\Rightarrow {{\left( z+xy \right)}^{2}}=\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right) \\ \Rightarrow {{z}^{2}}+{{x}^{2}}{{y}^{2}}+2xyz=1-{{x}^{2}}-{{y}^{2}}+{{x}^{2}}{{y}^{2}} \\
\therefore {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1 \\ \]
Hence, the correct option is B.

Note: We need to be careful about the domain of the inverse trigonometric ratios. Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $-\infty \le x\le \infty $. In that case we have to use the formula $x=n\pi \pm a$ for $\cos \left( x \right)=\cos a$ where $0\le a\le \pi $.