2. If $$ \sin \alpha-\frac{3}{5} $$ with $$ \alpha \in\left[90^{\circ}, 270^{\circ} ight] $$ and $$ \cos \beta=-\frac{12}{13} $$. wht $$ \beta \in\left[0^{\circ}, 10^{\circ} ight] $$ Calcutcte withot the use क a calculatr and with the aid of a diagram. te value \& $$ \cos \alpha+ an \beta $$.

Question

2. If $$ \sin \alpha-\frac{3}{5} $$ with $$ \alpha \in\left[90^{\circ}, 270^{\circ}ight] $$ and $$ \cos \beta=-\frac{12}{13} $$. wht $$ \beta \in\left[0^{\circ}, 10^{\circ}ight] $$ Calcutcte withot the use क a calculatr and with the aid of a diagram. te value \& $$ \cos \alpha+	an \beta $$.

UpStudy AI · Accepted Answer

To solve the problem, we need to find the values of $$ \alpha $$ and $$ \beta $$ based on the given conditions, and then calculate $$ \cos \alpha + \tan \beta $$.

### Step 1: Analyze $$ \alpha $$

We are given:
$$
\sin \alpha - \frac{3}{5} = 0
$$
This implies:
$$
\sin \alpha = \frac{3}{5}
$$

Since $$ \alpha $$ is in the interval $$ [90^{\circ}, 270^{\circ}] $$, we know that $$ \alpha $$ is in the second or third quadrant. In these quadrants, the sine function is positive, which is consistent with our value of $$ \sin \alpha $$.

To find $$ \alpha $$, we can use the reference angle:
$$
\alpha = 180^{\circ} - \theta \quad \text{(for the second quadrant)}
$$
or
$$
\alpha = 180^{\circ} + \theta \quad \text{(for the third quadrant)}
$$

Where $$ \theta $$ is the angle whose sine is $$ \frac{3}{5} $$. We can find $$ \theta $$ using the sine inverse function:
$$
\theta = \sin^{-1}\left(\frac{3}{5}\right)
$$

### Step 2: Calculate $$ \cos \alpha $$

Using the Pythagorean identity:
$$
\cos^2 \alpha + \sin^2 \alpha = 1
$$
we can find $$ \cos \alpha $$:
$$
\cos^2 \alpha = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}
$$
Thus,
$$
\cos \alpha = -\sqrt{\frac{16}{25}} = -\frac{4}{5} \quad \text{(since $$ \alpha $$ is in the second or third quadrant)}
$$

### Step 3: Analyze $$ \beta $$

We are given:
$$
\cos \beta = -\frac{12}{13}
$$
Since $$ \beta $$ is in the interval $$ [0^{\circ}, 10^{\circ}] $$, this means $$ \beta $$ must be in the first quadrant where cosine is positive. However, the given value of $$ \cos \beta $$ is negative, which indicates that $$ \beta $$ must actually be in the second quadrant.

To find $$ \beta $$, we can use the reference angle:
$$
\beta = 180^{\circ} - \phi
$$
where $$ \phi $$ is the angle whose cosine is $$ \frac{12}{13} $$.

### Step 4: Calculate $$ \tan \beta $$

Using the Pythagorean identity for sine:
$$
\sin^2 \beta + \cos^2 \beta = 1
$$
we can find $$ \sin \beta $$:
$$
\sin^2 \beta = 1 - \left(-\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169}
$$
Thus,
$$
\sin \beta = \frac{5}{13} \quad \text{(since $$ \beta $$ is in the second quadrant)}
$$

Now we can find $$ \tan \beta $$:
$$
\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{5}{13}}{-\frac{12}{13}} = -\frac{5}{12}
$$

### Step 5: Calculate $$ \cos \alpha + \tan \beta $$

Now we can calculate:
$$
\cos \alpha + \tan \beta = -\frac{4}{5} - \frac{5}{12}
$$

To add these fractions, we need a common denominator. The least common multiple of 5 and 12 is 60:
$$
-\frac{4}{5} = -\frac{48}{60}, \quad -\frac{5}{12} = -\frac{25}{60}
$$
Thus,
$$
\cos \alpha + \tan \beta = -\frac{48}{60} - \frac{25}{60} = -\frac{73}{60}
$$

### Final Answer

The value of $$ \cos \alpha + \tan \beta $$ is:
$$
\boxed{-\frac{73}{60}}
$$
The value of $$ \cos \alpha + \tan \beta $$ is $$ -\frac{73}{60} $$.

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