If `sintheta=(m^(2)-n^(2))/(m^(2)+n^(2))`, then `tantheta=?`

Options

A. `(m^(2)+n^(2))/(2mn)`
B. `(m^(2)-n^(2))/(2mn)`
C. `(m^(2)+n^(2))/(m^(2)-n^(2))`
D. `(m^(2)-n^(2))/(m^(2)+n^(2))`

Answer

To solve the problem where \( \sin \theta = \frac{m^2 - n^2}{m^2 + n^2} \) and we need to find \( \tan \theta \), we can follow these steps:

### Step 1: Use the Pythagorean Identity
We know from the Pythagorean identity that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
We will use this identity to find \( \cos^2 \theta \).

### Step 2: Calculate \( \sin^2 \theta \)
First, we calculate \( \sin^2 \theta \):
\[ \sin^2 \theta = \left(\frac{m^2 - n^2}{m^2 + n^2}\right)^2 = \frac{(m^2 - n^2)^2}{(m^2 + n^2)^2} \]

### Step 3: Substitute \( \sin^2 \theta \) into the Pythagorean Identity
Now we substitute \( \sin^2 \theta \) into the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
This gives us:
\[ \frac{(m^2 - n^2)^2}{(m^2 + n^2)^2} + \cos^2 \theta = 1 \]

### Step 4: Isolate \( \cos^2 \theta \)
Rearranging the equation, we find:
\[ \cos^2 \theta = 1 - \frac{(m^2 - n^2)^2}{(m^2 + n^2)^2} \]
To combine the terms, we can express 1 as:
\[ 1 = \frac{(m^2 + n^2)^2}{(m^2 + n^2)^2} \]
So we have:
\[ \cos^2 \theta = \frac{(m^2 + n^2)^2 - (m^2 - n^2)^2}{(m^2 + n^2)^2} \]

### Step 5: Simplify the Numerator
Using the difference of squares:
\[ a^2 - b^2 = (a - b)(a + b) \]
Let \( a = m^2 + n^2 \) and \( b = m^2 - n^2 \):
\[ (m^2 + n^2 - (m^2 - n^2))(m^2 + n^2 + (m^2 - n^2)) = (2n^2)(2m^2) = 4m^2n^2 \]
Thus, we have:
\[ \cos^2 \theta = \frac{4m^2n^2}{(m^2 + n^2)^2} \]

### Step 6: Calculate \( \cos \theta \)
Taking the square root gives us:
\[ \cos \theta = \frac{2mn}{m^2 + n^2} \]

### Step 7: Find \( \tan \theta \)
Now we can find \( \tan \theta \) using the definition:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{m^2 - n^2}{m^2 + n^2}}{\frac{2mn}{m^2 + n^2}} \]
This simplifies to:
\[ \tan \theta = \frac{m^2 - n^2}{2mn} \]

### Final Result
Thus, the value of \( \tan \theta \) is:
\[ \tan \theta = \frac{m^2 - n^2}{2mn} \]

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