1. **State the problem:** Simplify the expression $$(1+\cot A - \csc A)(1 + \tan A + \sec A)$$. 2. **Recall definitions:** - $\cot A = \frac{\cos A}{\sin A}$ - $\csc A = \frac{1}{\sin A}$ - $\tan A = \frac{\sin A}{\cos A}$ - $\sec A = \frac{1}{\cos A}$ 3. **Rewrite the expression using these:** $$\left(1 + \frac{\cos A}{\sin A} - \frac{1}{\sin A}\right)\left(1 + \frac{\sin A}{\cos A} + \frac{1}{\cos A}\right)$$ 4. **Simplify inside each parenthesis:** - First parentheses: $$1 + \frac{\cos A - 1}{\sin A} = \frac{\sin A}{\sin A} + \frac{\cos A - 1}{\sin A} = \frac{\sin A + \cos A - 1}{\sin A}$$ - Second parentheses: $$1 + \frac{\sin A + 1}{\cos A} = \frac{\cos A}{\cos A} + \frac{\sin A + 1}{\cos A} = \frac{\cos A + \sin A + 1}{\cos A}$$ 5. **Multiply the two simplified fractions:** $$\frac{\sin A + \cos A - 1}{\sin A} \times \frac{\cos A + \sin A + 1}{\cos A} = \frac{(\sin A + \cos A -1)(\sin A + \cos A + 1)}{\sin A \cos A}$$ 6. **Recognize the numerator as difference of squares:** $$(\sin A + \cos A)^2 - 1^2 = (\sin A + \cos A)^2 - 1$$ 7. **Expand $(\sin A + \cos A)^2$:** $$\sin^2 A + 2 \sin A \cos A + \cos^2 A$$ 8. **Use identity $\sin^2 A + \cos^2 A = 1$:** $$1 + 2 \sin A \cos A - 1 = 2 \sin A \cos A$$ 9. **Substitute back into numerator:** $$2 \sin A \cos A$$ 10. **Put together:** $$\frac{2 \sin A \cos A}{\sin A \cos A} = 2$$ **Final answer:** $$\boxed{2}$$