Question

lf $ta{n}^{-1}\left(\frac{x+1}{x-1}\right)+ta{n}^{-1}(\frac{x-1}{x})$ $=\pi +ta{n}^{-1}(-7)$, then $x=$

• A. $2$
• B. $-2$
• C. $1$
• D. $No$ $solution$

Answer 1

Answer: (A)

$2$

$ta{n}^{-1}x+ta{n}^{-1}y=\pi +ta{n}^{-1}\left(\frac{x+y}{1-xy}\right)$; $xy>1$ [formula]
$\therefore ta{n}^{-1}\left(\frac{x+1}{x-1}\right)+ta{n}^{-1}\left(\frac{x-1}{x}\right)=ta{n}^{-1}\left(\frac{\frac{x^2+x+x^2-2x+1}{x(x-1)}}{1-\frac{x+1}{x}}\right)+\pi$
$=ta{n}^{-1}\left(\frac{-2x^2+x-1)}{(x-1)}\right)+\pi$
as Given $ta{n}^{-1}(-7)=ta{n}^{-1}\left(\frac{-(2x^2-x+1)}{x-1}\right)$
$-7=-\left(\frac{2x^2-x+1}{x-1}\right)$
$7x-7=2x^2-x+1$
$2x^2-8x+8=0$

$x^2-4x+4=0$
so,

$x=2$