Inverse Trigonometric Functions Question 85
Question: The $ +ve $ integral solution of $ {{\tan }^{-1}}x+{{\cos }^{-1}}\frac{y}{\sqrt{1+y^{2}}}={{\sin }^{-1}}\frac{3}{\sqrt{10}} $ is
Options:
A) $ x=1,y=2;x=2,y=7 $
B) $ x=1,y=3;x=2,y=4 $
C) $ x=0,y=0;x=3,y=4 $
D) None of these
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Answer:
Correct Answer: A
Converting cos and sin into tan, we get, $ {{\tan }^{-1}}x+{{\tan }^{-1}}( \frac{1}{y} )={{\tan }^{-1}}( \frac{3}{1} ) $
$ \Rightarrow {{\tan }^{-1}}[ \frac{x+(1/y)}{1-x(1/y)} ]={{\tan }^{-1}}3 $
$ \Rightarrow {{\tan }^{-1}}( \frac{xy+1}{y-x} )={{\tan }^{-1}}3\Rightarrow \frac{xy+1}{y-x}=3 $
$ \Rightarrow x=\frac{3y-1}{y+3}=3-\frac{10}{y+3}\therefore <3 $ So, for $ x=1,y=2 $ and for $ z=2,y=7 $
BETA
