Question

If we have an inverse trigonometric expression ${{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi $ , then prove that ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1$

Answer 1

Hint: From the question we have ${{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi $ and we have to prove ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1$ using the given condition so we will assume ${{\cos }^{-1}}x=A$ and ${{\cos }^{-1}}y=B$ and simplify it further using the expressions $\pi -{{\cos }^{-1}}z={{\cos }^{-1}}\left( -z \right)$ and $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ and ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.

Complete step-by-step solution:
Now from considering the question we have ${{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi $. Let us ${{\cos }^{-1}}x=A$ and ${{\cos }^{-1}}y=B$ and write this simply as$A+B=\pi -{{\cos }^{-1}}z$ .
As we know that the value of $\pi -{{\cos }^{-1}}z={{\cos }^{-1}}\left( -z \right)$ we will use it and simplify it and write it as $A+B={{\cos }^{-1}}\left( -z \right)$.
By applying the inverse of cosine function on both sides we will have $\cos \left( A+B \right)=\left( -z \right)$ .
As we know that $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$. We can use this by deriving the values of respective sine functions using the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
Now we can say that $\sin A=\sqrt{1-{{x}^{2}}}$ and $\sin B=\sqrt{1-{{y}^{2}}}$ because $\cos A=x$ and $\cos B=y$ and they are connected with the relation${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ .
By using these values in $\cos \left( A+B \right)=\left( -z \right)$ we will have $xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}}=-z$ .
This expression can be simplified as $\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}}=xy+z$ .
By squaring on both sides we will have ${{\left( \sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right)}^{2}}={{\left( xy+z \right)}^{2}}\Rightarrow \left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)={{\left( xy+z \right)}^{2}}$ .
By expanding this we will have $\left( 1+{{x}^{2}}{{y}^{2}}-{{y}^{2}}-{{x}^{2}} \right)=\left( {{x}^{2}}{{y}^{2}}+{{z}^{2}}+2xyz \right)$ .
By further simplifying this we will have $\left( 1-{{y}^{2}}-{{x}^{2}} \right)=\left( {{z}^{2}}+2xyz \right)$ .
By transforming some terms from right to left and some from right to left we will have ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1$.
Hence it is proved that when ${{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi $ , the expression ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1$ is valid.

Note: While answering this type of questions we should keep 3 main points in our mind what we have, what is to be proved, and which concepts and formulae can be used if we make a mistake and write $\pi -{{\cos }^{-1}}z={{\cos }^{-1}}z$ then we will have $xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}}=z$ which by simplifying we will have ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-2xyz=1$ which is completely wrong.