A simpler way to prove $\frac{\tan A}{1-\cot A}+ \frac{\cot A}{1-\tan A}=\sec A \csc A + 1$?

Question

The question asks to prove: $$ \frac{\tan A}{1-\cot A}+ \frac{\cot A}{1-\tan A}=\sec A \csc A + 1$$ using only: $$ \sin^2A+\cos^2A=1\;\; \text{ & }\; \;\tan^2 A+1=\sec^2 A\;\; \text{ & } \;\; \cot^2 A+1=\csc^2 A $$

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My solution is join the two fractions, then multiplying numerator and denominator by $\tan^2 A$ to have only tangent functions: $$\frac{\tan A-\tan^2 A +\cot A-\cot^2 A}{ 2-\cot A-\tan A } \rightarrow \frac{\tan^3 A-\tan^4 A+\tan{A}-1}{2\tan^2 A -\tan A -\tan^3 A}$$ This then apparently factorises (I would never have spotted the top, it's only after trying it on Mathematica that I realised it did) to give:

$$ \frac{-(\tan A-1)^2(\tan^2 A+\tan A+1)}{-\tan A(\tan A-1)^2} =\cot A(\tan^2 A+\tan A+1)=1+\cot A+\tan A$$ which then simplifies down to the required result, but needless to say this isn't very elegant and not inline with the previous problems that whilst not easy didn't involve factoring a quartic which leads me to believe there is certainly an easier way to do it hence the question, I just can't seem to see it (I've done too many for today!).

Answer 1

I can't think of an "elegant" solution, but typically just converting everything to sin/cos is the way to go. For example, let $x=\sin(A)$ and $y=\cos(A)$. Then, $$ \frac{x/y}{1-y/x}+\frac{y/x}{1-x/y} = \frac{x^2/y}{x-y} - \frac{y^2/x}{x-y} = \frac{x^3-y^3}{xy(x-y)} = \frac{x^2+xy+y^2}{xy} = 1 + \frac{x^2+y^2}{xy}.$$

Now making use of the trig identity, we obtain simply $1+\frac{1}{xy}$, which is the result we want.

Answer 2

Prove the following simpler identities first, and they help build a more elegant solution:

Identity 1 $$\tan A+\cot A=\frac{\sin A}{cos A}+\frac{\cos A}{\sin A}=\frac{\sin^2A+\cos^2A}{\sin A\cos A}=\sec A\csc A$$

Identity 2 $$\sec^2A+\csc^2A=\frac{1}{\cos^2A}+\frac{1}{\sin^2A}=\frac{\sin^2A+\cos^2A}{\sin^2A\cos^2A}=\sec^2A\csc^2A$$

Identity 3 (using $\cot A=\frac{1}{\tan A}$ and Identity 1) $$(1-\cot A)(1-\tan A)=1-\cot A-\tan A+\cot A\tan A=2-\cot A-\tan A=2-\sec A\csc A$$

Identity 4 (using Idenity 3) $$\tan^2A+\cot^2A=(\sec^2A-1)+(\csc^2A-1)=\sec^2A+\csc^2A-2=\sec^2A\csc^2A-2$$

With these identities, the following simplification can be made: $$\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=\frac{(\tan A-\tan^2A)+(\cot A-\cot^2A)}{(1-\cot A)(1-\tan A)}=\frac{\tan A+\cot A-(\tan^2A+\cot^2A)}{2-\sec A\csc A}=\frac{2+\sec A\csc A-\sec^2A\csc^2A}{2-\sec A\csc A}=\frac{(2-\sec A\csc A)(1+\sec A\csc A)}{2-\sec A\csc A}=1+\sec A\csc A$$