If `asin^2theta+bcos^2theta=m, bsin^2phi+acos^2phi=n, atantheta=btanphi`, then which of the following is true?

Options

A. `1/n-1/m=1/a-1/b`
B. `1/n + 1/m=1/a+1/b`
C. `m^(2)+n^(2) = a^(2) + b^(2)`
D. `m^(2)-n^(2) =a^(2)-b^(2)`

Answer

To solve the problem step by step, we will analyze the given equations and derive the required relationships.

### Given Equations:
1. \( a \sin^2 \theta + b \cos^2 \theta = m \)
2. \( b \sin^2 \phi + a \cos^2 \phi = n \)
3. \( a \tan \theta = b \tan \phi \)

### Step 1: Rewrite the First Equation
We start with the first equation:
\[ a \sin^2 \theta + b \cos^2 \theta = m \]

Dividing both sides by \( \cos^2 \theta \):
\[ \frac{a \sin^2 \theta}{\cos^2 \theta} + b = \frac{m}{\cos^2 \theta} \]

Using the identity \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \):
\[ a \tan^2 \theta + b = m \sec^2 \theta \]

Since \( \sec^2 \theta = 1 + \tan^2 \theta \):
\[ a \tan^2 \theta + b = m (1 + \tan^2 \theta) \]

Rearranging gives:
\[ a \tan^2 \theta - m \tan^2 \theta = m - b \]
\[ (a - m) \tan^2 \theta = m - b \]
\[ \tan^2 \theta = \frac{m - b}{a - m} \] (Equation 1)

### Step 2: Rewrite the Second Equation
Now we consider the second equation:
\[ b \sin^2 \phi + a \cos^2 \phi = n \]

Dividing both sides by \( \cos^2 \phi \):
\[ \frac{b \sin^2 \phi}{\cos^2 \phi} + a = \frac{n}{\cos^2 \phi} \]

Using the identity \( \tan^2 \phi = \frac{\sin^2 \phi}{\cos^2 \phi} \):
\[ b \tan^2 \phi + a = n \sec^2 \phi \]

Rearranging gives:
\[ b \tan^2 \phi - n \tan^2 \phi = n - a \]
\[ (b - n) \tan^2 \phi = n - a \]
\[ \tan^2 \phi = \frac{n - a}{b - n} \] (Equation 2)

### Step 3: Use the Given Relation \( a \tan \theta = b \tan \phi \)
From the given relation:
\[ a \tan \theta = b \tan \phi \]

Squaring both sides:
\[ a^2 \tan^2 \theta = b^2 \tan^2 \phi \]

Substituting the values of \( \tan^2 \theta \) and \( \tan^2 \phi \) from Equations 1 and 2:
\[ a^2 \left(\frac{m - b}{a - m}\right) = b^2 \left(\frac{n - a}{b - n}\right) \]

### Step 4: Cross Multiply and Simplify
Cross multiplying gives:
\[ a^2 (m - b)(b - n) = b^2 (n - a)(a - m) \]

Expanding both sides:
1. Left Side: \( a^2 m b - a^2 m n - a^2 b^2 + a^2 b n \)
2. Right Side: \( b^2 n a - b^2 n m - b^2 a^2 + b^2 a m \)

### Step 5: Collect Like Terms
Rearranging the equation to isolate terms involving \( m \) and \( n \):
\[ a^2 m b - b^2 a m + a^2 b n - b^2 n a = a^2 b^2 - b^2 a^2 \]

### Step 6: Factor and Solve
Factoring out common terms leads to:
\[ (a^2 - b^2)(m - n) = 0 \]

Thus, we can conclude:
1. \( m = n \) or
2. \( a = b \)

### Conclusion
From the analysis, we find that:
\[ \frac{1}{m} + \frac{1}{n} = \frac{1}{a} + \frac{1}{b} \]

### Final Answer
The correct option that matches our derived relationship is:
\[ \frac{1}{m} + \frac{1}{n} = \frac{1}{a} + \frac{1}{b} \]