If `(cosalpha)/(cosbeta)=m and (cosalpha)/(sinbeta)=n`, find `(m^(2)+n^(2))cos^(2)beta`

Answer

To solve the problem, we start with the given equations:

1. \(\frac{\cos \alpha}{\cos \beta} = m\)
2. \(\frac{\cos \alpha}{\sin \beta} = n\)

We need to find the expression \( (m^2 + n^2) \cos^2 \beta \).

### Step 1: Express \( m^2 \) and \( n^2 \)

From the first equation, we can express \( m^2 \):
\[ m^2 = \left(\frac{\cos \alpha}{\cos \beta}\right)^2 = \frac{\cos^2 \alpha}{\cos^2 \beta} \]

From the second equation, we can express \( n^2 \):
\[ n^2 = \left(\frac{\cos \alpha}{\sin \beta}\right)^2 = \frac{\cos^2 \alpha}{\sin^2 \beta} \]

### Step 2: Substitute \( m^2 \) and \( n^2 \) into the expression

Now, substitute \( m^2 \) and \( n^2 \) into the expression \( (m^2 + n^2) \cos^2 \beta \):
\[ m^2 + n^2 = \frac{\cos^2 \alpha}{\cos^2 \beta} + \frac{\cos^2 \alpha}{\sin^2 \beta} \]

### Step 3: Combine the fractions

To combine the two fractions, we need a common denominator:
\[ m^2 + n^2 = \frac{\cos^2 \alpha \sin^2 \beta + \cos^2 \alpha \cos^2 \beta}{\cos^2 \beta \sin^2 \beta} \]
\[ = \frac{\cos^2 \alpha (\sin^2 \beta + \cos^2 \beta)}{\cos^2 \beta \sin^2 \beta} \]

Using the Pythagorean identity \(\sin^2 \beta + \cos^2 \beta = 1\):
\[ m^2 + n^2 = \frac{\cos^2 \alpha}{\cos^2 \beta \sin^2 \beta} \]

### Step 4: Multiply by \( \cos^2 \beta \)

Now we multiply the entire expression by \( \cos^2 \beta \):
\[ (m^2 + n^2) \cos^2 \beta = \frac{\cos^2 \alpha}{\sin^2 \beta} \]

### Step 5: Simplify using the definition of \( n \)

Since we know from the problem statement that \( n = \frac{\cos \alpha}{\sin \beta} \):
\[ (m^2 + n^2) \cos^2 \beta = n^2 \]

### Final Answer

Thus, the final result is:
\[ (m^2 + n^2) \cos^2 \beta = n^2 \]