Inverse Trigonometric Functions Question 191
Question: If $ {{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi $ , then
[Roorkee 1994]
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
Solution:
Given that $ {{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi $
$ \Rightarrow {{\cos }^{-1}}(x)+{{\cos }^{-1}}(y)+{{\cos }^{-1}}(z)={{\cos }^{-1}}(-1) $
$ \Rightarrow {{\cos }^{-1}}(x)+{{\cos }^{-1}}(y)={{\cos }^{-1}}(-1)-{{\cos }^{-1}}(z) $
$ \Rightarrow {{\cos }^{-1}}(xy-\sqrt{1-x^{2}}\sqrt{1-y^{2})}={{\cos }^{-1}}{ (-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 $ .
Trick: Put $ x=y=z=\frac{1}{2}, $ so that $ {{\cos }^{-1}}\frac{1}{2}+{{\cos }^{-1}}\frac{1}{2}+{{\cos }^{-1}}\frac{1}{2}=\pi $
Obviously (d) holds for these values of x, y, z.
BETA
