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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let } \mathrm{f}(x)=	an ^{-1} x \ 	herefore \quad & \mathrm{f}^{\prime}(x)=\frac{1}{1+x^2} \end{aligned}$ Here, $\mathrm{a}=1$ and $\mathrm{h}=-0.001$ $\begin{aligned} 	herefore \quad f(a+h) \approx f(a) & +h f^{\prime}(a) \ 	herefore \quad 	an ^{-1}(0.999) & \approx \frac{\pi}{4}+\frac{1}{1+1}(-0.001) \ & \approx \frac{\pi}{4}-\frac{0.001}{2} \ & \approx \frac{\pi}{4}-0.0005 \ & \approx \frac{3.1415}{4}-0.0005 \ & \approx 0.7849 \end{aligned}$

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

Solution