Prove that:$$\frac{\tan{x}}{1-\cot{x}} + \frac{\cot{x}}{1-\tan{x}}=\sec{x}\csc{x}+1.$$

My trying:$$L.H.S.=\frac{\tan{x}+\cot{x}-\tan^2{x}-\cot^2{x}}{(1-\cot{x})(1-\tan{x})}$$ Changing to $\sin{x}$ and $\cos{x}$ and simplifying gives (not typing the entire simplifying, its very lengthy) $$\frac{\sin{x}\cos{x}+2\sin{x}\cos{x}-1}{(\sin{x}\cos{x})(2\sin{x}\cos{x}-1)}$$ Tried many times but no luck getting anywhere close to the expression on R.H.S.


Since $a^3-b^3=(a-b)(a^2+ab+b^2)$ and $\sin^2x+\cos^2x=1$, we obtain:

$$\frac{\tan{x}}{1-\cot{x}} + \frac{\cot{x}}{1-\tan{x}}=\frac{\sin^2x}{\cos{x}(\sin{x}-\cos{x})}-\frac{\cos^2x}{\sin{x}(\sin{x}-\cos{x})}=$$ $$=\frac{\sin^3x-\cos^3x}{\sin{x}\cos{x}(\sin{x}-\cos{x})}=\frac{1+\sin{x}\cos{x}}{\sin{x}\cos{x}}=\sec{x}\csc{x}+1.$$

  • $\begingroup$ wow, I wasted like half an hour on this! $\endgroup$ – R3l1c Sep 3 '17 at 15:46

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