1. **Problem statement:** Prove the identity $$\frac{1+\tan x}{1+\cot x} \equiv \frac{1+\tan x}{\cot x -1}$$ 2. **Recall definitions:** - $\tan x = \frac{\sin x}{\cos x}$ - $\cot x = \frac{\cos x}{\sin x}$ 3. **Rewrite the left-hand side (LHS):** $$\frac{1+\tan x}{1+\cot x} = \frac{1 + \frac{\sin x}{\cos x}}{1 + \frac{\cos x}{\sin x}}$$ 4. **Find common denominators inside numerator and denominator:** Numerator: $$1 + \frac{\sin x}{\cos x} = \frac{\cos x}{\cos x} + \frac{\sin x}{\cos x} = \frac{\cos x + \sin x}{\cos x}$$ Denominator: $$1 + \frac{\cos x}{\sin x} = \frac{\sin x}{\sin x} + \frac{\cos x}{\sin x} = \frac{\sin x + \cos x}{\sin x}$$ 5. **Rewrite LHS as:** $$\frac{\frac{\cos x + \sin x}{\cos x}}{\frac{\sin x + \cos x}{\sin x}} = \frac{\cos x + \sin x}{\cos x} \times \frac{\sin x}{\sin x + \cos x}$$ 6. **Simplify numerator and denominator terms:** Since $\cos x + \sin x = \sin x + \cos x$, these cancel out: $$= \frac{\cancel{\cos x + \sin x}}{\cos x} \times \frac{\sin x}{\cancel{\sin x + \cos x}} = \frac{\sin x}{\cos x} = \tan x$$ 7. **Rewrite the right-hand side (RHS):** $$\frac{1+\tan x}{\cot x -1} = \frac{1 + \frac{\sin x}{\cos x}}{\frac{\cos x}{\sin x} - 1}$$ 8. **Find common denominators inside numerator and denominator:** Numerator: $$1 + \frac{\sin x}{\cos x} = \frac{\cos x + \sin x}{\cos x}$$ Denominator: $$\frac{\cos x}{\sin x} - 1 = \frac{\cos x}{\sin x} - \frac{\sin x}{\sin x} = \frac{\cos x - \sin x}{\sin x}$$ 9. **Rewrite RHS as:** $$\frac{\frac{\cos x + \sin x}{\cos x}}{\frac{\cos x - \sin x}{\sin x}} = \frac{\cos x + \sin x}{\cos x} \times \frac{\sin x}{\cos x - \sin x}$$ 10. **No immediate simplification to $\tan x$ here, so check if the original identity is correct:** The original identity claims: $$\frac{1+\tan x}{1+\cot x} \equiv \frac{1+\tan x}{\cot x -1}$$ But from step 6, LHS simplifies to $\tan x$, while RHS simplifies to $$\frac{\cos x + \sin x}{\cos x} \times \frac{\sin x}{\cos x - \sin x}$$ which is not equal to $\tan x$ in general. 11. **Conclusion:** The given identity is not true as stated. **Therefore, the identity $$\frac{1+\tan x}{1+\cot x} \equiv \frac{1+\tan x}{\cot x -1}$$ is false.**