Detailed solution to find α + β given tan α = m/(m+1) and tan β = 1/(2m+1).

Trigonometry Problem: Find \( \alpha + \beta \)

Hello, student! Let’s solve this problem together. You’re given:

\( \tan \alpha = \frac{m}{m + 1} \) and \( \tan \beta = \frac{1}{2m + 1} \)

Your task is to find the value of \( \alpha + \beta \). Here are the options:

a) \( \frac{\pi}{3} \)

b) \( \frac{\pi}{4} \)

c) \( \frac{\pi}{6} \)

d) none of the above

Detailed Solution

Let’s find \( \alpha + \beta \) step-by-step using the information provided. Since we have \( \tan \alpha \) and \( \tan \beta \), and we need the sum of the angles, we’ll use a special trigonometric formula. Don’t worry—I’ll explain everything clearly!

Step 1: Understand the Formula

To find \( \alpha + \beta \), we can use the tangent addition formula, which tells us what \( \tan(\alpha + \beta) \) is:

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

If we can calculate \( \tan(\alpha + \beta) \), we can figure out the angle \( \alpha + \beta \) by matching it to the tangent values we know (like \( \tan \frac{\pi}{4} = 1 \), \( \tan \frac{\pi}{3} = \sqrt{3} \), etc.).

Step 2: Substitute the Given Values

We’re given:

\( \tan \alpha = \frac{m}{m + 1} \)

\( \tan \beta = \frac{1}{2m + 1} \)

Let’s plug these into the formula:

\[ \tan(\alpha + \beta) = \frac{\frac{m}{m + 1} + \frac{1}{2m + 1}}{1 - \left( \frac{m}{m + 1} \right) \left( \frac{1}{2m + 1} \right)} \]

Step 3: Simplify the Numerator

First, let’s add the two fractions in the numerator: \( \frac{m}{m + 1} + \frac{1}{2m + 1} \).

The denominators are different, so we need a common denominator. The common denominator is \( (m + 1)(2m + 1) \). Rewrite each term:

\[ \frac{m}{m + 1} = \frac{m (2m + 1)}{(m + 1)(2m + 1)} \]

\[ \frac{1}{2m + 1} = \frac{1 (m + 1)}{(m + 1)(2m + 1)} \]

Now add them:

\[ \frac{m (2m + 1)}{(m + 1)(2m + 1)} + \frac{m + 1}{(m + 1)(2m + 1)} = \frac{m (2m + 1) + (m + 1)}{(m + 1)(2m + 1)} \]

Expand the numerator:

\[ m (2m + 1) + (m+1) = (2m^2 + m)+ (m+1) \]

\[ 2m^2 + m + m + 1 = 2m^2 + 2m + 1 \]

So, the numerator is:

\[ \frac{2m^2 + 2m + 1}{(m + 1)(2m + 1)} \]

Step 4: Simplify the Denominator

Now, tackle the denominator: \( 1 - \left( \frac{m}{m + 1} \right) \left( \frac{1}{2m + 1} \right) \).

First, multiply the fractions:

\[ \frac{m}{m + 1} \cdot \frac{1}{2m + 1} = \frac{m}{(m + 1)(2m + 1)} \]

So the denominator becomes:

\[ 1 - \frac{m}{(m + 1)(2m + 1)} \]

\[ \frac{(m + 1)(2m + 1)}{(m + 1)(2m + 1)} - \frac{m}{(m + 1)(2m + 1)} = \frac{(m + 1)(2m + 1) - m}{(m + 1)(2m + 1)} \]

Expand \( (m + 1)(2m + 1) \):

\[ (m + 1)(2m + 1) = m \cdot 2m + m \cdot 1 + 1 \cdot 2m + 1 \cdot 1 = 2m^2 + m + 2m + 1 = 2m^2 + 3m + 1 \]

Then:

\[ 2m^2 + 3m + 1 - m = 2m^2 + 2m + 1 \]

So, the denominator is:

\[ \frac{2m^2 + 2m + 1}{(m + 1)(2m + 1)} \]

Step 5: Put It Together

Now we have:

\[ \tan(\alpha + \beta) = \frac{\frac{2m^2 + 2m + 1}{(m + 1)(2m + 1)}}{\frac{2m^2 + 2m + 1}{(m + 1)(2m + 1)}} \]

Notice something amazing—the numerator and denominator are the same! When we divide a number by itself (as long as it’s not zero), we get 1:

\[ \tan(\alpha + \beta) = 1 \]

(Note: \( 2m^2 + 2m + 1 \) is always positive for real \( m \), and the denominator is defined except at \( m = -1 \) or \( m = -\frac{1}{2} \), which we assume are excluded.)

Step 6: Find the Angle

If \( \tan(\alpha + \beta) = 1 \), what is \( \alpha + \beta \)? We know:

\[ \tan \frac{\pi}{4} = 1 \]

Final Answer

So, \( \alpha + \beta = \frac{\pi}{4} \), which is option b).