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

& {{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi \\\ & \Rightarrow \theta +\beta +\alpha =\pi \\\ \end{aligned}$$ Now, taking $$\alpha $$ to the RHS of the above equation, we get, $$\begin{aligned} & \theta +\beta +\alpha =\pi \\\ & \Rightarrow \theta +\beta =\pi -\alpha \\\ \end{aligned}$$ Now, taking cosine in LHS and RHS, we get $$\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{aligned} & \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{aligned}$$ Now, squaring both sides, we get $$\begin{aligned} & \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{aligned}$$ So, LHS = RHS. Hence, proved. Note:This question can also be solved by solving the LHS of $${{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{aligned} & {{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{aligned}$$