`(cos(90^(@)-theta).sec(90^(@)-theta).tantheta)/(cosec(90^(@)-theta).sin(90^(@)-theta).cot(90^(@)-theta))=?`

Options

A. 1
B. 2
C. 3
D. `-1`

Answer

To solve the expression \((\cos(90^\circ - \theta) \cdot \sec(90^\circ - \theta) \cdot \tan \theta) / (\csc(90^\circ - \theta) \cdot \sin(90^\circ - \theta) \cdot \cot(90^\circ - \theta))\), we will use trigonometric identities.

### Step-by-step Solution:

1. **Use the co-function identities**:
- \(\cos(90^\circ - \theta) = \sin \theta\)
- \(\sec(90^\circ - \theta) = \csc \theta\)
- \(\csc(90^\circ - \theta) = \sec \theta\)
- \(\sin(90^\circ - \theta) = \cos \theta\)
- \(\cot(90^\circ - \theta) = \tan \theta\)

Substituting these identities into the expression gives:
\[ \frac{\sin \theta \cdot \csc \theta \cdot \tan \theta}{\sec \theta \cdot \cos \theta \cdot \tan \theta} \]

2. **Simplify the expression**:
- Recall that \(\csc \theta = \frac{1}{\sin \theta}\) and \(\sec \theta = \frac{1}{\cos \theta}\).
- Therefore, we can rewrite the expression:
\[ \frac{\sin \theta \cdot \frac{1}{\sin \theta} \cdot \tan \theta}{\frac{1}{\cos \theta} \cdot \cos \theta \cdot \tan \theta} \]

3. **Cancel out terms**:
- The \(\sin \theta\) in the numerator and denominator cancels out:
\[ \frac{1 \cdot \tan \theta}{\frac{1}{\cos \theta} \cdot \cos \theta \cdot \tan \theta} \]
- The \(\tan \theta\) in the numerator and denominator also cancels out:
\[ \frac{1}{1} = 1 \]

4. **Final Result**:
The final result of the expression is:
\[ 1 \]