Answer is: option4

1.05

Solution:

We are given:

\( y = f(x) = \sqrt{1 - \sin x} \).

At \( x = 0 \):

\( f(0) = \sqrt{1 - \sin 0} = \sqrt{1 - 0} = \sqrt{1} = 1 \).

Using the chain rule:

\( f(x) = (1 - \sin x)^{\frac{1}{2}} \).

Differentiate using the chain rule:

\( f'(x) = \frac{1}{2} (1 - \sin x)^{-\frac{1}{2}} (-\cos x) \).

\( f'(x) = \frac{-\cos x}{2\sqrt{1 - \sin x}} \).

At \( x = 0 \):

\( f'(0) = \frac{-\cos 0}{2\sqrt{1 - \sin 0}} = \frac{-1}{2 \cdot \sqrt{1}} = \frac{-1}{2} \).

The equation of the tangent line at \( x = 0 \) is:

\( y \approx f(0) + f'(0)(x - 0) \).

Substituting values:

\( y \approx 1 + \left(\frac{-1}{2}\right)(x) \).

For \( x = -0.1 \):

\( y \approx 1 + \left(\frac{-1}{2} \times (-0.1)\right) \).

\( y \approx 1 + \frac{0.1}{2} = 1 + 0.05 = 1.05 \).

The answer is 1.05.