Inverse Trigonometric Functions Question 59
Question: If $ {{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi , $ then
Options:
A) $ x^{2}+y^{2}+z^{2}+xyz=0 $
B) $ x^{2}+y^{2}+z^{2}+2xyz=0 $
C) $ x^{2}+y^{2}+z^{2}+xyz=1 $
D) $ x^{2}+y^{2}+z^{2}+2xyz=1 $
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Answer:
Correct Answer: D
Given that $ {{\cos }^{-1}}(x)+co{s^{-1}}(y)+co{s^{-1}}(z)=\pi $
$ \Rightarrow {{\cos }^{-1}}(x)+co{s^{-1}}(y)+co{s^{-1}}(z)={{\cos }^{-1}}(-1) $
$ \Rightarrow {{\cos }^{-1}}(x)+co{s^{-1}}(y)=\pi -co{s^{-1}}(z) $
$ \Rightarrow {{\cos }^{-1}}(xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}})=co{s^{-1}}(-z) $
$ \Rightarrow xy-\sqrt{(1-x^{2})(1-y^{2})}=-z $
$ \Rightarrow (xy+z)=\sqrt{(1-x^{2})(1-y^{2})} $
Squaring both sides, we get $ x^{2}+y^{2}+z^{2}+2xyz=1. $
BETA
