The approximate value of tan ^{-1}(0.999) is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $f(x) = 	an^{-1} x$. Then $f'(x) = \frac{1}{1+x^2}$. We use the linear approximation formula $f(a+h) \approx f(a) + h f'(a)$. Here,let $a = 1

Vedclass · Accepted Answer

$0.7849$

Vedclass · Answer

(B) Let $f(x) = 	an^{-1} x$. Then $f'(x) = \frac{1}{1+x^2}$. We use the linear approximation formula $f(a+h) \approx f(a) + h f'(a)$. Here,let $a = 1$ and $h = -0.001$,so $a+h = 0.999$. $f(a) = 	an^{-1}(1) = \frac{\pi}{4}$. $f'(a) = \frac{1}{1+1^2} = \frac{1}{2} = 0.5$. Substituting these values: $	an^{-1}(0.999) \approx \frac{\pi}{4} + (-0.001)(0.5)$. $	an^{-1}(0.999) \approx \frac{3.1415}{4} - 0.0005$. $	an^{-1}(0.999) \approx 0.785375 - 0.0005$. $	an^{-1}(0.999) \approx 0.784875$. Rounding to four decimal places,we get $0.7849$.

The approximate value of $\tan ^{-1}(0.999)$ is (use $\pi=3.1415$ )

Similar Questions

The approximate value of $\sqrt{63}$ is

By considering $1^{\prime} = 0.0175$ radians,the approximate value of $\cot 45^{\circ} 2^{\prime}$ is

Find the approximate change in the volume $V$ of a cube of side $x$ meters caused by increasing the side by $2 \%$.

The approximate value of $\frac{1}{(2.002)^2}$ is

The radius of a cone of height $9 \text{ units}$ is changed from $2 \text{ units}$ to $2.12 \text{ units}$. The exact change and approximate change in the volume of the cone are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module