If `l cos^(2) theta + m sin^(2) theta = (cos^(2) theta ("cosec"^(2) theta+1)/("cosec"^(2) theta -1))` then what is the value of `tan^(2) theta`.

Options

A. `(l-2)/(m-1)`
B. `(l-1)/(2-m)`
C. `(l-2)/(l-m)`
D. `(2-l)/(1-m)`

Answer

To solve the equation \( l \cos^2 \theta + m \sin^2 \theta = \frac{\cos^2 \theta \left( \csc^2 \theta + 1 \right)}{\csc^2 \theta - 1} \), we will follow these steps:

### Step 1: Rewrite the Right-Hand Side
We start with the right-hand side (RHS) of the equation:
\[ \frac{\cos^2 \theta \left( \csc^2 \theta + 1 \right)}{\csc^2 \theta - 1} \]
Recall that \( \csc^2 \theta = \frac{1}{\sin^2 \theta} \). Thus, we can rewrite the RHS as:
\[ \frac{\cos^2 \theta \left( \frac{1}{\sin^2 \theta} + 1 \right)}{\frac{1}{\sin^2 \theta} - 1} \]

### Step 2: Simplify the Expression
Now, simplify the numerator and denominator:
- The numerator becomes:
\[ \cos^2 \theta \left( \frac{1 + \sin^2 \theta}{\sin^2 \theta} \right) = \frac{\cos^2 \theta (1 + \sin^2 \theta)}{\sin^2 \theta} \]
- The denominator becomes:
\[ \frac{1 - \sin^2 \theta}{\sin^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} \]

### Step 3: Combine the Fractions
Thus, the RHS can be rewritten as:
\[ \frac{\frac{\cos^2 \theta (1 + \sin^2 \theta)}{\sin^2 \theta}}{\frac{\cos^2 \theta}{\sin^2 \theta}} = \frac{\cos^2 \theta (1 + \sin^2 \theta)}{\cos^2 \theta} \]
This simplifies to:
\[ 1 + \sin^2 \theta \]

### Step 4: Equate Both Sides
Now we equate both sides:
\[ l \cos^2 \theta + m \sin^2 \theta = 1 + \sin^2 \theta \]

### Step 5: Rearranging the Equation
Rearranging gives us:
\[ l \cos^2 \theta + m \sin^2 \theta - \sin^2 \theta = 1 \]
This can be expressed as:
\[ l \cos^2 \theta + (m - 1) \sin^2 \theta = 1 \]

### Step 6: Divide by \( \cos^2 \theta \)
Now, divide the entire equation by \( \cos^2 \theta \):
\[ l + (m - 1) \tan^2 \theta = \sec^2 \theta \]
Using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \), we can rewrite this as:
\[ l + (m - 1) \tan^2 \theta = 1 + \tan^2 \theta \]

### Step 7: Rearranging for \( \tan^2 \theta \)
Rearranging gives:
\[ (m - 1) \tan^2 \theta - \tan^2 \theta = 1 - l \]
Factoring out \( \tan^2 \theta \):
\[ (m - 2) \tan^2 \theta = 1 - l \]

### Step 8: Solve for \( \tan^2 \theta \)
Finally, we can solve for \( \tan^2 \theta \):
\[ \tan^2 \theta = \frac{1 - l}{m - 2} \]

### Final Answer
Thus, the value of \( \tan^2 \theta \) is:
\[ \tan^2 \theta = \frac{l - 1}{2 - m} \]

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