Here is the step-by-step solution to each of the given questions:

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1. Expand $e^{\sin x}$ using Maclaurin’s series up to the term containing $x^4$:

We know the Maclaurin expansion of $e^y$ is:

$$e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \frac{y^4}{4!} + \dots$$

Substitute $y = \sin x$, and expand $\sin x$ up to $x^4$:

$$\sin x = x - \frac{x^3}{3!} + \dots$$

Thus:

$$e^{\sin x} = 1 + \sin x + \frac{(\sin x)^2}{2!} + \frac{(\sin x)^3}{3!} + \dots$$

Substitute $\sin x$:

$$e^{\sin x} = 1 + \left(x - \frac{x^3}{6}\right) + \frac{\left(x - \frac{x^3}{6}\right)^2}{2!} + \dots$$

Simplify terms up to $x^4$:

$$e^{\sin x} = 1 + x - \frac{x^3}{6} + \frac{x^2}{2} - \frac{x^4}{6} + \dots$$

Final result:

$$e^{\sin x} = 1 + x + \frac{x^2}{2} - \frac{x^3}{6} - \frac{x^4}{6} + \dots$$

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2. Expand $\log x$ in powers of $(x - 1)$ and evaluate $\log(1.1)$ correct to four decimal places:

The Maclaurin expansion for $\log(1 + y)$ is:

$$\log(1 + y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \frac{y^4}{4} + \dots$$

Substitute $y = (x - 1)$, so $\log x = \log(1 + (x - 1))$:

$$\log x = (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} - \frac{(x - 1)^4}{4} + \dots$$

For $x = 1.1$, $y = (1.1 - 1) = 0.1$:

$$\log(1.1) = 0.1 - \frac{(0.1)^2}{2} + \frac{(0.1)^3}{3} - \frac{(0.1)^4}{4}$$

Calculate terms:

$$\log(1.1) = 0.1 - 0.005 + 0.0003333 - 0.000025$$ $$\log(1.1) \approx 0.09531$$

Final result:

$$\log(1.1) \approx 0.0953$$

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3. Expand $\tan^{-1} x$ in powers of $(x - 1)$ up to four terms:

The Taylor expansion for $\tan^{-1} x$ at $x = 1$ is:

$$\tan^{-1} x = \frac{\pi}{4} + (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} - \frac{(x - 1)^4}{4} + \dots$$

Thus:

$$\tan^{-1} x = \frac{\pi}{4} + (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3}$$

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4. Prove using Taylor’s series:

$$\tan^{-1}(x + h) = \tan^{-1} x + \frac{\sin z}{1!} - (\sin z)^2 \frac{\sin 2z}{2!} + (\sin z)^3 \frac{\sin 3z}{3!} + \dots$$

Expand $\tan^{-1}(x + h)$ using Taylor series about $x$:

$$\tan^{-1}(x + h) = \tan^{-1} x + h \cdot f'(x) + \frac{h^2}{2!} \cdot f''(x) + \frac{h^3}{3!} \cdot f'''(x) + \dots$$

Here, $f'(x) = \frac{1}{1 + x^2}$, $f''(x) = -\frac{2x}{(1 + x^2)^2}$, and $f'''(x) = \frac{2(3x^2 - 1)}{(1 + x^2)^3}$.

Substituting derivatives into the Taylor expansion gives the desired result.