Step-by-Step Solution
Key Concept: Use the fundamental trigonometric definitions $\sec A = \frac{1}{\cos A}$, $\tan A = \frac{\sin A}{\cos A}$ and the Pythagorean identity $1-\sin^{2}A = \cos^{2}A$ to simplify the expression.
1. Write the given expression:
$$E = \sec A\,(1-\sin A)\,(\sec A+\tan A).$$
2. Replace $\sec A$ and $\tan A$ by their definitions in terms of sine and cosine:
$$\sec A = \frac{1}{\cos A}, \qquad \tan A = \frac{\sin A}{\cos A}.$$
Thus
$$\sec A + \tan A = \frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}.$$
3. Substitute this back into $E$:
$$E = \frac{1}{\cos A}\,(1-\sin A)\,\frac{1+\sin A}{\cos A}.
$$
4. Multiply the two fractions:
$$E = \frac{(1-\sin A)(1+\sin A)}{\cos^{2}A}.
$$
5. Use the difference of squares identity $ (1-\sin A)(1+\sin A)=1-\sin^{2}A$:
$$E = \frac{1-\sin^{2}A}{\cos^{2}A}.$$
6. Apply the Pythagorean identity $1-\sin^{2}A = \cos^{2}A$:
$$E = \frac{\cos^{2}A}{\cos^{2}A}=1.$$
7. Hence, the given expression simplifies to $1$, proving the identity.
Therefore, $\boxed{\sec A\,(1-\sin A)(\sec A+\tan A)=1}$.
Correct Answer: 1