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Hint: So we have to find the approximate value ${{\tan }^{-1}}(0.999)$.Use $a=1$ and $h=0.001$. Find $f(a)$and${{f}^{'}}(a)$ substitute it in $f(a+h)\approx f(a)+h{{f}^{'}}(a)$. You will get the answer.

Complete step-by-step answer:

The linear function whose graph is the tangent line to $y=f(x)$ at a specified point $(a,f(a))$ is called the linear approximation of $f(x)$ near $x=a$.

When a physical measurement is made, there is always some uncertainty about its accuracy. For instance, if you are measuring the radius of a ball bearing, you might measure it repeatedly and obtain slightly differing results.

Rather than concluding, say, that the radius of the ball bearing is exactly $1.2mm$, you may instead conclude that the radius is $1.2mm\pm 0.1mm$.

(The actual calculation of the range $\pm 0.1$ is often given by a statistical formula based on the standard deviation of all the separate measurements.)

Once you have an error estimate for the radius, you might wonder how this error might affect the calculation of the volume of the ball bearing. In other words, if the radius is off by $0.1mm$,

by how much is the volume off? To answer the question, think of the error of the radius as a change,

$\Delta r$, in $r$, and then compute the associated change, $\Delta V$, in the volume $V$.

Linear approximations of $f(x)$near$x=a$.

It $x$ is close to $a$, then $f(x)\approx f(a)+(x-a){{f}^{'}}(a)$.

So now we are given the question that we have to find an approximate value ${{\tan }^{-1}}(0.999)$.

Let us take it $f(x)={{\tan }^{-1}}(x)$.

So we know the differentiation of ${{\tan }^{-1}}x$ which is given by,

${{f}^{'}}(x)=\dfrac{1}{1+{{x}^{2}}}$

Now let us assume that it lets us take $a=1$ and $h=0.001$.

So substituting above values we get,

$f(a)=f(1)={{\tan }^{-1}}(1)=\dfrac{\pi }{4}$ radians

And ${{f}^{'}}(a)={{f}^{'}}(1)=\dfrac{1}{1+{{1}^{2}}}=\dfrac{1}{2}=0.5$ radians.

The approximation formula is given below.

$f(a+h)\approx f(a)+h{{f}^{'}}(a)$

So now substituting the values in this approximation formula, we get,

${{\tan }^{-1}}(0.999)\approx f(1)+(0.001){{f}^{'}}(1)$

So now simplifying in a simple manner we get,

${{\tan }^{-1}}(0.999)\approx \dfrac{\pi }{4}+(0.001)(0.5)$

Again simplifying we get,

${{\tan }^{-1}}(0.999)\approx \dfrac{\pi }{4}+0.0005\approx 0.7855$

So making the number in the simple form we get,

So we get,

${{\tan }^{-1}}(0.999)\approx 0.787$

So we get the correct answer as an option(C).

Note: Read the question carefully. You should be familiar with the process of solving the approximations. You should know the general form that is $f(a+h)\approx f(a)+h{{f}^{'}}(a)$.

Don’t jumble yourself while differentiating and substituting.

Complete step-by-step answer:

The linear function whose graph is the tangent line to $y=f(x)$ at a specified point $(a,f(a))$ is called the linear approximation of $f(x)$ near $x=a$.

When a physical measurement is made, there is always some uncertainty about its accuracy. For instance, if you are measuring the radius of a ball bearing, you might measure it repeatedly and obtain slightly differing results.

Rather than concluding, say, that the radius of the ball bearing is exactly $1.2mm$, you may instead conclude that the radius is $1.2mm\pm 0.1mm$.

(The actual calculation of the range $\pm 0.1$ is often given by a statistical formula based on the standard deviation of all the separate measurements.)

Once you have an error estimate for the radius, you might wonder how this error might affect the calculation of the volume of the ball bearing. In other words, if the radius is off by $0.1mm$,

by how much is the volume off? To answer the question, think of the error of the radius as a change,

$\Delta r$, in $r$, and then compute the associated change, $\Delta V$, in the volume $V$.

Linear approximations of $f(x)$near$x=a$.

It $x$ is close to $a$, then $f(x)\approx f(a)+(x-a){{f}^{'}}(a)$.

So now we are given the question that we have to find an approximate value ${{\tan }^{-1}}(0.999)$.

Let us take it $f(x)={{\tan }^{-1}}(x)$.

So we know the differentiation of ${{\tan }^{-1}}x$ which is given by,

${{f}^{'}}(x)=\dfrac{1}{1+{{x}^{2}}}$

Now let us assume that it lets us take $a=1$ and $h=0.001$.

So substituting above values we get,

$f(a)=f(1)={{\tan }^{-1}}(1)=\dfrac{\pi }{4}$ radians

And ${{f}^{'}}(a)={{f}^{'}}(1)=\dfrac{1}{1+{{1}^{2}}}=\dfrac{1}{2}=0.5$ radians.

The approximation formula is given below.

$f(a+h)\approx f(a)+h{{f}^{'}}(a)$

So now substituting the values in this approximation formula, we get,

${{\tan }^{-1}}(0.999)\approx f(1)+(0.001){{f}^{'}}(1)$

So now simplifying in a simple manner we get,

${{\tan }^{-1}}(0.999)\approx \dfrac{\pi }{4}+(0.001)(0.5)$

Again simplifying we get,

${{\tan }^{-1}}(0.999)\approx \dfrac{\pi }{4}+0.0005\approx 0.7855$

So making the number in the simple form we get,

So we get,

${{\tan }^{-1}}(0.999)\approx 0.787$

So we get the correct answer as an option(C).

Note: Read the question carefully. You should be familiar with the process of solving the approximations. You should know the general form that is $f(a+h)\approx f(a)+h{{f}^{'}}(a)$.

Don’t jumble yourself while differentiating and substituting.

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