1. **State the problem:** Prove that $$1 + \tan A + \sec A = \frac{2}{1 + \cot A - \csc A}$$. 2. **Recall the definitions:** - $$\tan A = \frac{\sin A}{\cos A}$$ - $$\sec A = \frac{1}{\cos A}$$ - $$\cot A = \frac{\cos A}{\sin A}$$ - $$\csc A = \frac{1}{\sin A}$$ 3. **Rewrite the left side (LHS):** $$1 + \tan A + \sec A = 1 + \frac{\sin A}{\cos A} + \frac{1}{\cos A}$$ 4. **Find common denominator for LHS terms with denominator $$\cos A$$:** $$= \frac{\cos A}{\cos A} + \frac{\sin A}{\cos A} + \frac{1}{\cos A} = \frac{\cos A + \sin A + 1}{\cos A}$$ 5. **Rewrite the right side (RHS):** $$\frac{2}{1 + \cot A - \csc A} = \frac{2}{1 + \frac{\cos A}{\sin A} - \frac{1}{\sin A}}$$ 6. **Find common denominator $$\sin A$$ in the denominator of RHS:** $$= \frac{2}{\frac{\sin A}{\sin A} + \frac{\cos A}{\sin A} - \frac{1}{\sin A}} = \frac{2}{\frac{\sin A + \cos A - 1}{\sin A}}$$ 7. **Simplify RHS by dividing by a fraction:** $$= 2 \times \frac{\sin A}{\sin A + \cos A - 1} = \frac{2 \sin A}{\sin A + \cos A - 1}$$ 8. **Now we want to prove:** $$\frac{\cos A + \sin A + 1}{\cos A} = \frac{2 \sin A}{\sin A + \cos A - 1}$$ 9. **Cross multiply:** $$ (\cos A + \sin A + 1)(\sin A + \cos A - 1) = 2 \sin A \cos A $$ 10. **Expand the left side:** $$ (\cos A)(\sin A) + (\cos A)(\cos A) - (\cos A)(1) + (\sin A)(\sin A) + (\sin A)(\cos A) - (\sin A)(1) + 1 \times \sin A + 1 \times \cos A - 1 \times 1 $$ 11. **Simplify terms:** $$ \cos A \sin A + \cos^2 A - \cos A + \sin^2 A + \sin A \cos A - \sin A + \sin A + \cos A - 1 $$ 12. **Combine like terms:** - $$\cos A \sin A + \sin A \cos A = 2 \sin A \cos A$$ - $$-\cos A + \cos A = 0$$ - $$-\sin A + \sin A = 0$$ - $$\cos^2 A + \sin^2 A = 1$$ So the expression becomes: $$ 2 \sin A \cos A + 1 - 1 = 2 \sin A \cos A $$ 13. **Left side equals right side:** $$ 2 \sin A \cos A = 2 \sin A \cos A $$ 14. **Conclusion:** The original equation is proven true. **Final answer:** $$1 + \tan A + \sec A = \frac{2}{1 + \cot A - \csc A}$$.