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)

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