1. **State the problem:** Prove that $$\frac{\sin x}{1-\cos x} + \frac{\tan x}{1+\cos x} = \cot x + \sec x \csc x$$. 2. **Recall formulas and identities:** - $$\tan x = \frac{\sin x}{\cos x}$$ - $$\cot x = \frac{\cos x}{\sin x}$$ - $$\sec x = \frac{1}{\cos x}$$ - $$\csc x = \frac{1}{\sin x}$$ - Use the Pythagorean identity $$\sin^2 x + \cos^2 x = 1$$. 3. **Simplify the left-hand side (LHS):** $$\frac{\sin x}{1-\cos x} + \frac{\tan x}{1+\cos x} = \frac{\sin x}{1-\cos x} + \frac{\frac{\sin x}{\cos x}}{1+\cos x} = \frac{\sin x}{1-\cos x} + \frac{\sin x}{\cos x (1+\cos x)}$$ 4. **Find common denominator for LHS:** The denominators are $$1-\cos x$$ and $$\cos x (1+\cos x)$$. Multiply numerator and denominator of first term by $$\cos x (1+\cos x)$$ and second term by $$1-\cos x$$: $$\frac{\sin x \cos x (1+\cos x)}{(1-\cos x) \cos x (1+\cos x)} + \frac{\sin x (1-\cos x)}{\cos x (1+\cos x)(1-\cos x)}$$ 5. **Combine terms:** Denominator is $$\cos x (1-\cos x)(1+\cos x) = \cos x (1 - \cos^2 x) = \cos x \sin^2 x$$. Numerator: $$\sin x \cos x (1+\cos x) + \sin x (1-\cos x) = \sin x [\cos x (1+\cos x) + (1-\cos x)]$$ 6. **Simplify numerator inside brackets:** $$\cos x (1+\cos x) + (1-\cos x) = \cos x + \cos^2 x + 1 - \cos x = 1 + \cos^2 x$$ So numerator is $$\sin x (1 + \cos^2 x)$$. 7. **LHS becomes:** $$\frac{\sin x (1 + \cos^2 x)}{\cos x \sin^2 x} = \frac{1 + \cos^2 x}{\cos x \sin x}$$ 8. **Rewrite numerator:** $$1 + \cos^2 x = (\sin^2 x + \cos^2 x) + \cos^2 x = \sin^2 x + 2 \cos^2 x$$ So LHS = $$\frac{\sin^2 x + 2 \cos^2 x}{\cos x \sin x} = \frac{\sin^2 x}{\cos x \sin x} + \frac{2 \cos^2 x}{\cos x \sin x} = \frac{\sin x}{\cos x} + \frac{2 \cos x}{\sin x} = \tan x + 2 \cot x$$ 9. **Simplify right-hand side (RHS):** $$\cot x + \sec x \csc x = \frac{\cos x}{\sin x} + \frac{1}{\cos x} \cdot \frac{1}{\sin x} = \frac{\cos x}{\sin x} + \frac{1}{\cos x \sin x} = \cot x + \frac{1}{\cos x \sin x}$$ 10. **Compare LHS and RHS:** LHS = $$\tan x + 2 \cot x$$ RHS = $$\cot x + \frac{1}{\cos x \sin x}$$ Since $$\tan x = \frac{\sin x}{\cos x}$$ and $$\cot x = \frac{\cos x}{\sin x}$$, rewrite LHS: $$\tan x + 2 \cot x = \frac{\sin x}{\cos x} + 2 \frac{\cos x}{\sin x} = \frac{\sin^2 x}{\cos x \sin x} + \frac{2 \cos^2 x}{\cos x \sin x} = \frac{\sin^2 x + 2 \cos^2 x}{\cos x \sin x}$$ This matches the earlier expression for LHS. 11. **Check if RHS equals this:** RHS = $$\cot x + \sec x \csc x = \frac{\cos x}{\sin x} + \frac{1}{\cos x \sin x} = \frac{\cos^2 x + 1}{\cos x \sin x}$$ Recall $$\cos^2 x + 1 = 1 + \cos^2 x$$ which matches numerator of LHS. 12. **Conclusion:** LHS = RHS, so the identity is proven. **Final answer:** $$\boxed{\frac{\sin x}{1-\cos x} + \frac{\tan x}{1+\cos x} = \cot x + \sec x \csc x}$$