If `cosectheta-sintheta=landsectheta-costheta=m`, then `l^(2)m^(2)(l^(2)+m^(2)+3)=?`

Options

A. 1
B. `-1`
C. 2
D. 4

Answer

To solve the problem given by the equations \( \csc \theta - \sin \theta = l \) and \( \sec \theta - \cos \theta = m \), we need to find the value of \( l^2 m^2 (l^2 + m^2 + 3) \).

### Step 1: Express \( l \) and \( m \) in terms of \( \sin \theta \) and \( \cos \theta \)

Given:
\[ l = \csc \theta - \sin \theta = \frac{1}{\sin \theta} - \sin \theta \]
\[ m = \sec \theta - \cos \theta = \frac{1}{\cos \theta} - \cos \theta \]

### Step 2: Simplify \( l \)

Substituting for \( l \):
\[ l = \frac{1 - \sin^2 \theta}{\sin \theta} = \frac{\cos^2 \theta}{\sin \theta} \]

### Step 3: Simplify \( m \)

Substituting for \( m \):
\[ m = \frac{1 - \cos^2 \theta}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta} \]

### Step 4: Calculate \( l^2 \) and \( m^2 \)

Now, we calculate \( l^2 \) and \( m^2 \):
\[ l^2 = \left(\frac{\cos^2 \theta}{\sin \theta}\right)^2 = \frac{\cos^4 \theta}{\sin^2 \theta} \]
\[ m^2 = \left(\frac{\sin^2 \theta}{\cos \theta}\right)^2 = \frac{\sin^4 \theta}{\cos^2 \theta} \]

### Step 5: Calculate \( l^2 m^2 \)

Now we find \( l^2 m^2 \):
\[ l^2 m^2 = \left(\frac{\cos^4 \theta}{\sin^2 \theta}\right) \left(\frac{\sin^4 \theta}{\cos^2 \theta}\right) = \frac{\cos^4 \theta \sin^4 \theta}{\sin^2 \theta \cos^2 \theta} = \cos^2 \theta \sin^2 \theta \]

### Step 6: Calculate \( l^2 + m^2 \)

Next, we calculate \( l^2 + m^2 \):
\[ l^2 + m^2 = \frac{\cos^4 \theta}{\sin^2 \theta} + \frac{\sin^4 \theta}{\cos^2 \theta} \]
Finding a common denominator:
\[ = \frac{\cos^6 \theta + \sin^6 \theta}{\sin^2 \theta \cos^2 \theta} \]

Using the identity \( \sin^6 \theta + \cos^6 \theta = ( \sin^2 \theta + \cos^2 \theta)(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) = 1(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) \):
\[ = \frac{1 - 3 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \]

### Step 7: Combine to find \( l^2 m^2 (l^2 + m^2 + 3) \)

Now we can find \( l^2 m^2 (l^2 + m^2 + 3) \):
\[ l^2 + m^2 + 3 = \frac{1 - 3 \sin^2 \theta \cos^2 \theta + 3 \sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta} \]
Thus,
\[ l^2 m^2 (l^2 + m^2 + 3) = \cos^2 \theta \sin^2 \theta \cdot \frac{1}{\sin^2 \theta \cos^2 \theta} = 1 \]

### Final Answer:
\[ l^2 m^2 (l^2 + m^2 + 3) = 1 \]