Question

If \[{{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi \] , then prove that \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1\] .

Answer 1

Hint: Assume, \[\theta ={{\cos }^{-1}}x\] , \[\beta ={{\cos }^{-1}}y\] , and \[\alpha ={{\cos }^{-1}}z\] . Using this, transform the equation
\[{{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi \] . Solve the equation \[\theta +\beta =\pi -\alpha \] , using the property \[cos(\pi -\alpha )=-cos\alpha \] and the formula \[\cos (\theta +\beta )=cos\theta cos\beta -sin\theta sin\beta \] . Then using the identity \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \Rightarrow \sin \theta =\sqrt{1-{{\cos }^{2}}\theta }\] , get the values of \[\sin \theta \] and \[\sin \beta \] . Then put the values of \[\cos \theta \] , \[\cos \beta \] , \[\cos \alpha \] , \[\sin \theta \] and \[\sin \beta \] in the equation,
 \[cos\theta cos\beta -sin\theta sin\beta =-\cos \alpha \] and solve it further.

Complete step-by-step answer:
According to the question, it is given that,
\[{{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi \] …………………………(1)
Let us assume,
\[\theta ={{\cos }^{-1}}x\] …………………(2)
\[\beta ={{\cos }^{-1}}y\] …………………….(3)
\[\alpha ={{\cos }^{-1}}z\] ………………..(4)
Now, using equation (2), equation (3), and equation (4), we can transform equation (1).
On transforming equation (1), we get
\[\begin{align}
  & {{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi \\
 & \Rightarrow \theta +\beta +\alpha =\pi \\
\end{align}\]
Now, taking \[\alpha \] to the RHS of the above equation, we get,
\[\begin{align}
  & \theta +\beta +\alpha =\pi \\
 & \Rightarrow \theta +\beta =\pi -\alpha \\
\end{align}\]
Now, taking cosine in LHS and RHS, we get
\[\Rightarrow \theta +\beta =\pi -\alpha \]
\[\Rightarrow \cos (\theta +\beta )=cos(\pi -\alpha )\] ……………………(5)
We know the property, \[cos(\pi -\alpha )=-cos\alpha \] ……………………..(6)
We also know the formula, \[\cos (\theta +\beta )=cos\theta cos\beta -sin\theta sin\beta \] ……………………….(7)
Using equation (6) and equation (7), we can transform equation (5).
On transforming we get
\[\Rightarrow \cos (\theta +\beta )=cos(\pi -\alpha )\]
\[\Rightarrow cos\theta cos\beta -sin\theta sin\beta =-\cos \alpha \] …………………………(8)
To solve the above equation, we need the values of \[\sin \theta \] and \[\sin \beta \] . We don’t have values of \[\sin \theta \] and \[\sin \beta \] . So, we have to find the values of \[\sin \theta \] and \[\sin \beta \] .
We know the identity, \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \Rightarrow \sin \theta =\sqrt{1-{{\cos }^{2}}\theta }\].
Putting the value of \[\cos \theta \] from equation (1) in the above identity, we get
\[\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }=\sqrt{1-{{x}^{2}}}\] ……………………….(9)
Replacing \[\theta \] by \[\beta \] in the identity, we get
\[{{\cos }^{2}}\beta +{{\sin }^{2}}\beta =1\Rightarrow {{\sin }^{2}}\beta =1-{{\cos }^{2}}\beta \Rightarrow \sin \beta =\sqrt{1-{{\cos }^{2}}\beta }\]
Putting the value of \[\cos \beta \] from equation (2) in the above identity, we get
\[\sin \beta =\sqrt{1-{{\cos }^{2}}\beta }=\sqrt{1-{{y}^{2}}}\] ……………………….(10)
From equation (8), equation (9), and equation (10), we get,
\[\begin{align}
  & \Rightarrow cos\theta cos\beta -sin\theta sin\beta =-\cos \alpha \\
 & \Rightarrow xy-\sqrt{1-{{x}^{2}}}.\sqrt{1-{{y}^{2}}}=-z \\
 & \Rightarrow xy+z=\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \\
\end{align}\]
Now, squaring both sides, we get
\[\begin{align}
  & \Rightarrow {{(xy+z)}^{2}}={{\left( \sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \right)}^{2}} \\
 & \Rightarrow {{x}^{2}}{{y}^{2}}+{{z}^{2}}+2xyz=\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right) \\
 & \Rightarrow {{x}^{2}}{{y}^{2}}+{{z}^{2}}+2xyz=1-{{y}^{2}}-{{x}^{2}}+{{x}^{2}}{{y}^{2}} \\
 & \Rightarrow {{y}^{2}}+{{x}^{2}}+{{z}^{2}}+2xyz=1 \\
\end{align}\]
So, LHS = RHS.
Hence, proved.

Note:This question can also be solved by solving the LHS of \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1\] and then making it equal to 1. For that, just pout the value of x, y, and z in the above equation and then solve it further.
\[\begin{align}
  & {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz \\
 & ={{\cos }^{2}}\theta +{{\cos }^{2}}\beta +{{\cos }^{2}}\alpha +2\cos \theta \cos \beta \cos \alpha \\
 & ={{x}^{2}}+{{y}^{2}}+{{\left( \sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)}-xy \right)}^{2}}+2xy\left( \sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)}-xy \right) \\
 & ={{x}^{2}}+{{y}^{2}}+\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)+{{x}^{2}}{{y}^{2}}-2xy\left( \sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \right)+2xy\left( \sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)}-xy \right) \\
 & ={{x}^{2}}+{{y}^{2}}+1-{{x}^{2}}-{{y}^{2}}+{{x}^{2}}{{y}^{2}}+{{x}^{2}}{{y}^{2}}-2xy\left( \sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \right)+2xy\left( \sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \right)-2{{x}^{2}}{{y}^{2}} \\
 & =1+2{{x}^{2}}{{y}^{2}}-2{{x}^{2}}{{y}^{2}} \\
 & =1 \\
\end{align}\]