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