已知 $\sin A = 0.3$，其中 $A$ 为钝角。
求下列三角比的值。

1. Find the value of $\cos A$ using the identity $\sin^2 A + \cos^2 A = 1$ and the given quadrant.

Given $\sin A = 0.3 = \frac{3}{10}$.
$\cos^2 A = 1 - \sin^2 A = 1 - (0.3)^2 = 1 - 0.09 = 0.91 = \frac{91}{100}$.
Since $A$ is obtuse ($90^\circ < A < 180^\circ$), $\cos A$ is negative.
$\cos A = -\sqrt{\frac{91}{100}} = -\frac{\sqrt{91}}{10}$.

2. Calculate $\sec A$ as the reciprocal of $\cos A$.

$\sec A = \frac{1}{\cos A} = \frac{1}{-\frac{\sqrt{91}}{10}} = -\frac{10}{\sqrt{91}}$.

1. Apply the double angle formula for cosine in terms of sine

$\cos 2A = 1 - 2\sin^2 A$

2. Substitute the given value of $\sin A$ and evaluate

$\cos 2A = 1 - 2(0.3)^2$
$\cos 2A = 1 - 2(0.09)$
$\cos 2A = 1 - 0.18 = 0.82$

1. Find the exact value of $\tan A$

Since $\sin A = 0.3 = \frac{3}{10}$ and $A$ is obtuse, $\cos A$ is negative.
$\cos A = -\sqrt{1 - 0.3^2} = -\sqrt{0.91} = -\frac{\sqrt{91}}{10}$
$\tan A = \frac{\sin A}{\cos A} = \frac{\frac{3}{10}}{-\frac{\sqrt{91}}{10}} = -\frac{3}{\sqrt{91}}$

2. Apply the addition formula for tangent

$\tan(A + 45^\circ) = \frac{\tan A + \tan 45^\circ}{1 - \tan A \tan 45^\circ}$

3. Substitute the values and simplify the expression

Using $\tan 45^\circ = 1$:
$\tan(A + 45^\circ) = \frac{-\frac{3}{\sqrt{91}} + 1}{1 - \left(-\frac{3}{\sqrt{91}}\right)(1)} = \frac{\frac{\sqrt{91} - 3}{\sqrt{91}}}{\frac{\sqrt{91} + 3}{\sqrt{91}}} = \frac{\sqrt{91} - 3}{\sqrt{91} + 3}$

4. Rationalise the denominator

$\frac{\sqrt{91} - 3}{\sqrt{91} + 3} \times \frac{\sqrt{91} - 3}{\sqrt{91} - 3} = \frac{91 - 6\sqrt{91} + 9}{91 - 9}$
$= \frac{100 - 6\sqrt{91}}{82}$
$= \frac{50 - 3\sqrt{91}}{41}$