Question

If $1^o=0.01745$ radians. Then the approximate value of $\tan {46}^0$ is

• A. $1.0259$
• B. $1.0394$
• C. $1.0349$
• D. $1.0493$

Answer 1

The correct option is C $1.0349$

Solution:- (C) $1.0349$

Given:- $1°=0.01745\text{radians}$

To find:- $\tan 46°=?$

Solution:-

$\tan 46°=\tan 45+1$

$\Rightarrow =\frac{\tan 45°+\tan 1°}{1-\tan 45°\tan 1°}$

$\Rightarrow =\tan \frac{\pi }{4}+\tan 1°[\because 1-\tan 45°\tan 1°\approx 1]$

Let $y=f\left(x\right)=\tan x$

$x=45°=\frac{\pi }{4}$

$\Delta x=1°=0.01745\text{rad.}$

Therefore,

$y=f\left(\frac{\pi }{4}\right)=\tan \frac{\pi }{4}=1$

Let $dx=\Delta x=1°=0.01745\text{rad}$

$y=\tan x$

Differentiating w.r.t. $x$, we have

$\frac{dy}{dx}={\sec }^{2}x$

at $x=\frac{\pi }{4}\Rightarrow \frac{dy}{dx}={\sec }^2\frac{\pi }{4}=2$

$\therefore dy=\frac{dy}{dx}\times dx=2\times 0.01745=0.0349\text{rad.}[\because dx=\Delta x=0.01745]$

$\therefore \tan 46°=y+\Delta y=1+0.0349=1.0349$