If `l cos^2 theta + m sin^2 theta =(cos^2 theta (cosec^2theta +1))/(cosec^2 theta -1), 0^@ lt theta lt 90^@` , then ` tan theta` is equal to

Options

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

Answer

To solve the equation given in the problem, we will follow these steps:

### Step 1: Rewrite the given equation
The equation provided is:

\[ l \cos^2 \theta + m \sin^2 \theta = \frac{\cos^2 \theta (\csc^2 \theta + 1)}{\csc^2 \theta - 1} \]

### Step 2: Simplify the right-hand side
We know that:

\[ \csc^2 \theta = \frac{1}{\sin^2 \theta} \]

Thus, we can rewrite the right-hand side:

\[ \csc^2 \theta + 1 = \frac{1}{\sin^2 \theta} + 1 = \frac{1 + \sin^2 \theta}{\sin^2 \theta} \]

And:

\[ \csc^2 \theta - 1 = \frac{1}{\sin^2 \theta} - 1 = \frac{1 - \sin^2 \theta}{\sin^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} \]

Substituting these into the right-hand side gives:

\[ \frac{\cos^2 \theta \left(\frac{1 + \sin^2 \theta}{\sin^2 \theta}\right)}{\frac{\cos^2 \theta}{\sin^2 \theta}} = \frac{\cos^2 \theta (1 + \sin^2 \theta)}{\cos^2 \theta} = 1 + \sin^2 \theta \]

### Step 3: Substitute back into the equation
Now, we substitute this back into the original equation:

\[ l \cos^2 \theta + m \sin^2 \theta = 1 + \sin^2 \theta \]

### Step 4: Rearranging the equation
Rearranging gives:

\[ l \cos^2 \theta + m \sin^2 \theta - \sin^2 \theta = 1 \]

This simplifies to:

\[ l \cos^2 \theta + (m - 1) \sin^2 \theta = 1 \]

### Step 5: Express in terms of tangent
Using the identity \(\sin^2 \theta = \frac{\tan^2 \theta}{1 + \tan^2 \theta}\) and \(\cos^2 \theta = \frac{1}{1 + \tan^2 \theta}\), we can substitute these into the equation:

\[ l \frac{1}{1 + \tan^2 \theta} + (m - 1) \frac{\tan^2 \theta}{1 + \tan^2 \theta} = 1 \]

### Step 6: Clear the denominator
Multiply through by \(1 + \tan^2 \theta\):

\[ l + (m - 1) \tan^2 \theta = 1 + \tan^2 \theta \]

### Step 7: Rearranging to find \(\tan^2 \theta\)
Rearranging gives:

\[ (m - 1) \tan^2 \theta - \tan^2 \theta = 1 - l \]

This simplifies to:

\[ ((m - 1) - 1) \tan^2 \theta = 1 - l \]

Thus:

\[ (m - 2) \tan^2 \theta = 1 - l \]

### Step 8: Solve for \(\tan^2 \theta\)
Finally, we can express \(\tan^2 \theta\):

\[ \tan^2 \theta = \frac{1 - l}{m - 2} \]

### Step 9: Take the square root to find \(\tan \theta\)
Taking the square root:

\[ \tan \theta = \sqrt{\frac{1 - l}{m - 2}} \]

### Final Answer
Thus, the value of \(\tan \theta\) is:

\[ \tan \theta = \sqrt{\frac{1 - l}{m - 2}} \]