For `0^(@) lt theta lt 90^(@)` which of the following expression is of `theta` ?
(i) `costheta(1- sin theta)^(-1) + cos theta (1 + sintheta)^(-1)`
(ii) `cos theta (1+ "cosec"theta)^(-1) + cos theta ("cosec"theta -1)^(-1)`
Choose the correct code among following.

Options

A. Only (i)
B. Only (ii)
C. Both (i) and (ii)
D. Neither (i) Nor (ii)

Answer

To solve the problem, we will analyze both expressions step by step.

### Expression (i):
\[ \text{Expression 1: } \cos \theta (1 - \sin \theta)^{-1} + \cos \theta (1 + \sin \theta)^{-1} \]

1. **Rewrite the expression**:
\[ \cos \theta \cdot \frac{1}{1 - \sin \theta} + \cos \theta \cdot \frac{1}{1 + \sin \theta} \]

2. **Factor out \(\cos \theta\)**:
\[ \cos \theta \left( \frac{1}{1 - \sin \theta} + \frac{1}{1 + \sin \theta} \right) \]

3. **Find a common denominator**:
The common denominator for the fractions is \((1 - \sin \theta)(1 + \sin \theta)\):
\[ = \cos \theta \left( \frac{(1 + \sin \theta) + (1 - \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)} \right) \]

4. **Simplify the numerator**:
\[ = \cos \theta \left( \frac{2}{1 - \sin^2 \theta} \right) \]

5. **Use the identity \(1 - \sin^2 \theta = \cos^2 \theta\)**:
\[ = \cos \theta \cdot \frac{2}{\cos^2 \theta} \]

6. **Simplify**:
\[ = \frac{2 \cos \theta}{\cos^2 \theta} = \frac{2}{\cos \theta} = 2 \sec \theta \]

### Expression (ii):
\[ \text{Expression 2: } \cos \theta (1 + \csc \theta)^{-1} + \cos \theta (\csc \theta - 1)^{-1} \]

1. **Rewrite the expression**:
\[ \cos \theta \cdot \frac{1}{1 + \csc \theta} + \cos \theta \cdot \frac{1}{\csc \theta - 1} \]

2. **Factor out \(\cos \theta\)**:
\[ \cos \theta \left( \frac{1}{1 + \csc \theta} + \frac{1}{\csc \theta - 1} \right) \]

3. **Find a common denominator**:
The common denominator is \((1 + \csc \theta)(\csc \theta - 1)\):
\[ = \cos \theta \left( \frac{(\csc \theta - 1) + (1 + \csc \theta)}{(1 + \csc \theta)(\csc \theta - 1)} \right) \]

4. **Simplify the numerator**:
\[ = \cos \theta \left( \frac{2 \csc \theta}{(1 + \csc \theta)(\csc \theta - 1)} \right) \]

5. **Rewrite \(\csc \theta\) as \(\frac{1}{\sin \theta}\)**:
\[ = \cos \theta \cdot \frac{2 \cdot \frac{1}{\sin \theta}}{(1 + \frac{1}{\sin \theta})(\frac{1}{\sin \theta} - 1)} \]

6. **Simplify the denominator**:
\[ = \cos \theta \cdot \frac{2}{\sin \theta} \cdot \frac{\sin^2 \theta}{(1 + 1)(1 - \sin \theta)} = \frac{2 \cos \theta \sin \theta}{\sin^2 \theta - 1} \]

7. **Use the identity \(\sin^2 \theta - 1 = -\cos^2 \theta\)**:
\[ = \frac{-2 \cos \theta \sin \theta}{\cos^2 \theta} = -2 \tan \theta \]

### Conclusion:
Both expressions contain \(\theta\):
1. Expression (i) simplifies to \(2 \sec \theta\).
2. Expression (ii) simplifies to \(-2 \tan \theta\).

Thus, the correct answer is that both expressions involve \(\theta\).

### Final Answer:
Both expressions have \(\theta\), so the answer is option C.

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