Inverse Trigonometric Functions Question 194
Question: If $ {{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\frac{\pi }{2}, $ then
[Karnataka CET 1996]
Options:
A) $ x+y+z-xyz=0 $
B) $ x+y+z+xyz=0 $
C) $ xy+yz+zx+1=0 $
D) $ xy+yz+zx-1=0 $
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Answer:
Correct Answer: D
Solution:
Given that $ {{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\frac{\pi }{2} $
$ \Rightarrow {{\tan }^{-1}}[ \frac{x+y+z-xyz}{1-xy-yz-xz} ]=\frac{\pi }{2} $
$ \Rightarrow [ \frac{x+y+z-xyz}{1-xy-yz-zx} ]=\tan \frac{\pi }{2}=\frac{1}{0} $
Hence $ xy+yz+zx-1=0 $ .
Trick : $ x=y=z=\frac{1}{\sqrt{3}}, $
so that $ {{\tan }^{-1}}\frac{1}{\sqrt{3}}+{{\tan }^{-1}}\frac{1}{\sqrt{3}}+{{\tan }^{-1}}\frac{1}{\sqrt{3}}=\frac{\pi }{2} $
Obviously (d) holds for these values of x, y, z.
BETA
