Question

If $x=\frac{\sqrt{3}}{3}$ and $(x+1)(y+1)=2$, then the value of ${\tan }^{-1}y$ is equal to

• A. $\frac{\pi }{12}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (A)

$x=\frac{1}{\sqrt{3}}$ and $xy+x+y+1=2$
$xy+x+y=1$
$y(1+x)=1-x$
$y=\frac{1-x}{1+x}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\Rightarrow y=\tan \frac{\pi }{12}$
$\therefore {\tan }^{-1}y=\frac{\pi }{12}$