Question

If ${\tan }^{-1}x+{\cos }^{-1}\frac{y}{\sqrt{1+y^2}}={\sin }^{-1}\frac{3}{\sqrt{10}}$ where $x,y\in N,$ then $x+y$ can be

• A. $3$
• B. $5$
• C. $9$
• D. $10$

Answer 1

Answer: (A), (C)

$9$

${\tan }^{-1}x+{\cos }^{-1}\frac{y}{\sqrt{1+y^2}}={\sin }^{-1}\frac{3}{\sqrt{10}}$
$\Rightarrow {\tan }^{-1}x+{\tan }^{-1}\left(\frac{1}{y}\right)={\tan }^{-1}3$
$\Rightarrow {\tan }^{-1}\left(\frac{1}{y}\right)={\tan }^{-1}3-{\tan }^{-1}x$
$\Rightarrow {\tan }^{-1}\left(\frac{1}{y}\right)={\tan }^{-1}\left(\frac{3-x}{1+3x}\right)$
$\Rightarrow y=\frac{1+3x}{3-x}$
As $x,y\in N,$
$x=1,2$ and correspondingly $y=2,7$
$\Rightarrow (x,y)=(1,2)$ or $(2,7)$
$\therefore x+y=3$ or $9$