Which of the folloiwng is true?`
a. cos thetasin theta-(sin theta cos (90^(@)-theta)cos theta)/(sec (90^(@)-theta))`
`-(cos theta sin (90^(@)-theta)sin theta)/(cosec(90^(@)-theta))=0`
b. If A and B are complementary angles, then `sinA=sqrt((cosA)/(sinB)-cosAsinB)`

Options

A. Only a
B. Only b
C. Neither a nor b
D. Both a and b

Answer

To determine which of the given statements is true, we will analyze both parts step by step.

### Part A:
We need to simplify the expression:
\[ \cos \theta \sin \theta - \frac{\sin \theta \cos(90^\circ - \theta) \cos \theta}{\sec(90^\circ - \theta)} - \frac{\cos \theta \sin(90^\circ - \theta) \sin \theta}{\csc(90^\circ - \theta)} = 0 \]

1. **Substituting Trigonometric Identities**:
- We know that \(\cos(90^\circ - \theta) = \sin \theta\) and \(\sin(90^\circ - \theta) = \cos \theta\).
- Also, \(\sec(90^\circ - \theta) = \frac{1}{\cos(90^\circ - \theta)} = \frac{1}{\sin \theta}\) and \(\csc(90^\circ - \theta) = \frac{1}{\sin(90^\circ - \theta)} = \frac{1}{\cos \theta}\).

Substituting these into the expression gives:
\[ \cos \theta \sin \theta - \frac{\sin \theta \sin \theta \cos \theta}{\frac{1}{\sin \theta}} - \frac{\cos \theta \cos \theta \sin \theta}{\frac{1}{\cos \theta}} = 0 \]

2. **Simplifying the Expression**:
- This simplifies to:
\[ \cos \theta \sin \theta - \sin^2 \theta \cos \theta - \cos^2 \theta \sin \theta = 0 \]

3. **Factoring the Expression**:
- We can factor out \(\sin \theta \cos \theta\):
\[ \sin \theta \cos \theta (1 - \sin \theta - \cos \theta) = 0 \]

4. **Using the Pythagorean Identity**:
- We know that \(\sin^2 \theta + \cos^2 \theta = 1\), thus \(1 - \sin^2 \theta - \cos^2 \theta = 0\).

5. **Conclusion for Part A**:
- Since \(\sin \theta \cos \theta\) can be zero or \(1 - \sin^2 \theta - \cos^2 \theta = 0\), the expression equals zero. Thus, **Part A is true**.

### Part B:
We need to evaluate the statement:
\[ \sin A = \sqrt{\frac{\cos A}{\sin B} - \cos A \sin B} \]
where \(A\) and \(B\) are complementary angles (\(A + B = 90^\circ\)).

1. **Using Complementary Angle Identities**:
- Since \(B = 90^\circ - A\), we have:
\[ \sin B = \cos A \]

2. **Substituting into the Expression**:
- Substitute \(\sin B\) into the equation:
\[ \sin A = \sqrt{\frac{\cos A}{\cos A} - \cos A \cos A} \]
- This simplifies to:
\[ \sin A = \sqrt{1 - \cos^2 A} \]

3. **Using the Pythagorean Identity**:
- We know that \(1 - \cos^2 A = \sin^2 A\), thus:
\[ \sin A = \sqrt{\sin^2 A} \]
- This implies \(\sin A = |\sin A|\). Since \(\sin A\) can be negative depending on the angle, the equality does not hold for all angles.

4. **Conclusion for Part B**:
- Thus, **Part B is false**.

### Final Conclusion:
- The true statement is **Part A**.