Question

If $\tan ^{-1} x+\cos ^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right)=\tan ^{-1} 4$ where $x, y \in N$, then the number of possible values of $x$, is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

$\tan ^{-1}\left(\frac{1}{y}\right)=\tan ^{-1}\left(\frac{4-x}{1+4 x}\right) $
$\therefore y=\frac{4 x+1}{4-x} \Rightarrow 4 y-x y=4 x+1 \Rightarrow x(y+4)=4 y-1 \Rightarrow x=\frac{4 y-1}{y+4}$
$x=\frac{4(y+4-4)-1}{y+4}=4-\frac{17}{y+4}$
$y+4=1 \Rightarrow y=-3 \text { (Rejected) } $
$y+4=17 \Rightarrow y=13, x=3$
$\therefore 1$