Let's solve this trigonometry problem involving the sum of two angles, $\alpha$ and $\beta$, given their tangent values.

We are given:

• $\tan \alpha = \frac{1}{2}$
• $\tan \beta = \frac{1}{3}$

We need to find the value of $\alpha + \beta$. To do this, we can use the tangent addition formula, which is a fundamental identity in trigonometry.

Applying the Tangent Addition Formula

The tangent addition formula for two angles, $\alpha$ and $\beta$, is given by:

$$ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} $$

This formula is very useful when you know the tangents of individual angles and need to find the tangent of their sum.

Step-by-Step Calculation of $\tan(\alpha + \beta)$

Now, let's substitute the given values of $\tan \alpha$ and $\tan \beta$ into the formula:

$$ \tan(\alpha + \beta) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \left(\frac{1}{2}\right) \left(\frac{1}{3}\right)} $$

First, calculate the sum in the numerator:

$$ \frac{1}{2} + \frac{1}{3} = \frac{1 \times 3}{2 \times 3} + \frac{1 \times 2}{3 \times 2} = \frac{3}{6} + \frac{2}{6} = \frac{3+2}{6} = \frac{5}{6} $$

Next, calculate the product in the denominator:

$$ \left(\frac{1}{2}\right) \left(\frac{1}{3}\right) = \frac{1 \times 1}{2 \times 3} = \frac{1}{6} $$

Now substitute these back into the tangent addition formula:

$$ \tan(\alpha + \beta) = \frac{\frac{5}{6}}{1 - \frac{1}{6}} $$

Calculate the difference in the denominator:

$$ 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{6-1}{6} = \frac{5}{6} $$

So, the expression becomes:

$$ \tan(\alpha + \beta) = \frac{\frac{5}{6}}{\frac{5}{6}} $$

Finally, simplify the fraction:

$$ \tan(\alpha + \beta) = 1 $$

Finding the Angle $\alpha + \beta$

We found that $\tan(\alpha + \beta) = 1$. Now we need to find the angle whose tangent is 1. We know from standard trigonometric values that the tangent of 45 degrees is 1.

$$ \tan(45^\circ) = 1 $$

Therefore, comparing this with our result, we can conclude that:

$$ \alpha + \beta = 45^\circ $$

This value matches one of the given options.

Analyzing the Options for $\alpha + \beta$

Let's quickly look at the options provided:

Our calculated value $\alpha + \beta = 45^\circ$ directly corresponds to Option 2.

Summary of Finding $\alpha + \beta$

By using the tangent addition formula and substituting the given values for $\tan \alpha$ and $\tan \beta$, we calculated $\tan(\alpha + \beta) = 1$. Recognizing that $\tan(45^\circ) = 1$, we determined that $\alpha + \beta$ equals $45^\circ$. This demonstrates the application of trigonometric identities to find the sum of angles.

Revision Table: Key Trigonometric Identities

Additional Information: Understanding Angles from Tangent Values

When finding an angle from its tangent value, like finding $\theta$ when $\tan \theta = k$, we are essentially using the inverse tangent function, written as $\arctan(k)$ or $\tan^{-1}(k)$.

• For $\tan(\alpha + \beta) = 1$, we have $\alpha + \beta = \arctan(1)$.
• The principal value of $\arctan(1)$ is $45^\circ$ (or $\pi/4$ radians).
• However, the tangent function has a period of $180^\circ$ (or $\pi$ radians). This means $\tan(\theta) = \tan(\theta + 180^\circ n)$ for any integer $n$.
• So, the general solution for $\tan(\theta) = 1$ is $\theta = 45^\circ + 180^\circ n$.
• In the context of finding the sum of two angles $\alpha$ and $\beta$, if $\alpha$ and $\beta$ are assumed to be acute angles (since $\tan \alpha$ and $\tan \beta$ are positive), then their sum $\alpha + \beta$ would typically fall within the range where the principal value of the inverse tangent is appropriate, which is $-90^\circ < \alpha + \beta < 90^\circ$. Since $\tan \alpha = 1/2$ and $\tan \beta = 1/3$ are positive, $\alpha$ and $\beta$ are likely in the first quadrant ($0^\circ$ to $90^\circ$). The sum of two angles in the first quadrant is between $0^\circ$ and $180^\circ$. Since $\tan(\alpha+\beta) = 1$ is positive, $\alpha+\beta$ must be in the first quadrant ($0^\circ$ to $90^\circ$). Thus, $45^\circ$ is the correct value for $\alpha + \beta$.

This problem is a classic example demonstrating how to use the sum of angles identity for the tangent function in trigonometry.