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...sigh..another problem how shall I prove the following? $$ {\cot A\over1- \tan A} + {\tan A \over 1- \cot A} = 1 + \tan A + \cot A$$

so what now? the following's what I've done: $$\cot A - \cot^2 A + \tan A- \tan^2 A \over 2 - \tan A - \cot A$$

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try to avoid LCM for the denominator before simplification. Observe that $1-\cot A=-\frac{1-\tan A}{\tan A}$ – lab bhattacharjee May 8 '13 at 15:12
up vote 5 down vote accepted

Let $\tan A=a\implies \cot A=\frac1a$

So, the problem reduces to

$$\frac{\frac1a}{1-a}+\frac a{1-\frac1a}=\frac1{a(1-a)}+\frac{a^2}{a-1}$$ $$=\frac1{a(1-a)}-\frac{a^2}{1-a}=\frac{1-a^3}{a(1-a)}=\frac{1+a+a^2}a=a+\frac1a+1$$


$$\frac{\tan A}{1-\cot A}=\frac{\tan^2A}{\tan A-1} (\text{ multiplying the numerator & the denominator by }\tan A) $$

$$\implies \frac{\tan A}{1-\cot A}=-\frac{\tan^2A}{1-\tan A}$$

$$\text{So,} {\cot A\over1- \tan A} + {\tan A \over 1- \cot A}={\cot A\over1- \tan A} -\frac{\tan^2A}{1-\tan A}=\frac{\cot A-\tan^2A}{1-\tan A}=\frac{1-\tan^3A}{\tan A(1-\tan A)} (\text{ multiplying the numerator & the denominator by }\tan A) $$

$$=\frac{1+\tan A+\tan^2A}{\tan A}\text{ (assuming }1-\tan A\ne0)$$

$$=\cot A+1+\tan A$$

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+1 for the first method that makes the problem clearer. (Also, don't you mean "assuming $\tan A ≠ 0$"?) – Mohammad Ali Baydoun May 8 '13 at 15:21
@Magtheridon96, no. Have you seen that I've cancelled $1-\tan A?$ – lab bhattacharjee May 8 '13 at 15:24
Oh I see. That assumption was referring to the previous step :v – Mohammad Ali Baydoun May 8 '13 at 15:25
How do you get $1 + tanA + tan^2A$ from $1-tan^3A$ in the last 2 steps? – Ghost May 8 '13 at 15:50
@Ghost, $a^3-b^3=(a-b)(a^2+ab+b^2)$ – lab bhattacharjee May 8 '13 at 15:51

$$ \begin{align} \frac{\cot A}{1-\tan A}+\frac{\tan A}{1-\cot A}&=\frac{\cos A\cot A - \sin A \tan A}{\cos A-\sin A} \\\\&=\frac{\frac{\cos^2 A}{\sin A} - \frac{\sin^2 A}{\cos A}}{\cos A-\sin A} \\\\&=\frac{\cos^3 A - \sin^3 A}{(\cos A \sin A)(\cos A-\sin A)} \\\\&=\frac{\cos^2 A + \sin^2 A + \sin A\cos A}{\cos A \sin A} \\\\&=1+\cot A+\tan A \end{align} $$

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