Question

The approximate value of ${\tan }^{-1}(0.999)$ (upto 4 decimal places ) is

• A. $0.7852$
• B. $0.7102$
• C. $0.8127$
• D. $0.7526$

Answer 1

Answer: (A)

Let $y={\tan }^{-1}x$
$y+\Delta y={\tan }^{-1}(x+\Delta x)$
$x=1,\Delta x=-0.001$
$\frac{dy}{dx}=\frac{1}{1+x^2}$
$dy=\frac{dx}{1+x^2}$
$dy=\frac{-0.001}{1+1}[\because dx\approx \Delta x]$
$dy=\frac{-0.001}{2}$
$\therefore {\tan }^{-1}(0.999)=y+\Delta y$
$={\tan }^{-1}1-\frac{0.001}{2}=\frac{\pi }{4}-\frac{0.001}{2}$
$=0.7857-0.0005=0.7852$