Question

Using differentials find the approximate value of $\tan\, 46^{\circ}$, if it is being given that $1^{\circ = 0.01745$ radians.

• A. $1.09$
• B. $2.01$
• C. $1.03$
• D. $1.001$

Answer 1

Answer: (C)

Let $y=f\left(x\right) = tan\,x, x=45^{\circ} = \left(\frac{\pi}{4}\right)^{c}$
and $x+\Delta x=46^{\circ}$.
Then,
$\Delta x = 1^{\circ} = 0.01745$ radians
For $x = \frac{\pi }{4}$, we have $y = f\left(\frac{\pi }{4}\right) = tan \frac{\pi }{4} = 1$
Let $dx = \Delta x = 0.01745$
Now, $y = tan \,x$
$\Rightarrow \frac{dy}{dx} = sec^{2}\,x$
$\Rightarrow \left(\frac{dy}{dx}\right)_{x=\pi/4} = sec^{2} \frac{\pi}{4} = 2$
$\therefore dy = \frac{dy}{dx}dx$
$\Rightarrow dy = 2\left(0.01745\right) = 0.03490$
$\Rightarrow \Delta y = 0.03490$
$\left(\because \Delta y \cong dy\right)$
Hence, $tan\, 46^{\circ} = y + \Delta y$
$= 1 + 0.03490$
$= 1.03490 \approx 1.03.$