Inverse Trigonometric Functions Question 62
Question: The equation $ {{\tan }^{-1}}(1+x)+ta{n^{-1}}(1-x)=\frac{\pi }{2} $ is satisfied by
Options:
A) $ x=1 $
B) $ x=-1 $
C) $ x=0 $
D) $ x=\frac{1}{2} $
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Answer:
Correct Answer: C
$ {{\tan }^{-1}}(1+x)+ta{n^{-1}}(1-x)=\frac{\pi }{2} $
$ {{\tan }^{-1}}[ \frac{(1+x)+(1-x)}{1-(1+x)(1-x)} ]=\frac{\pi }{2} $
$ \Rightarrow \frac{1+x+1-x}{1-(1+x)(1-x)}=\tan \frac{\pi }{2} $
$ \Rightarrow \frac{2}{1-(1+x)(1-x)}=\frac{1}{0}\Rightarrow 1-(1+x)(1-x)=0 $
$ \Rightarrow (1+x)(1-x)=1 $
$ 1-x^{2}=1 $
$ x^{2}=0 $ x = 0
BETA
