LHS $= \dfrac{\tan A}{1-\cot A} + \dfrac{\cot A}{1-\tan A}$
Convert to $\sin A$ and $\cos A$: let $s = \sin A$, $c = \cos A$.
$$= \frac{\tfrac{s}{c}}{1-\tfrac{c}{s}} + \frac{\tfrac{c}{s}}{1-\tfrac{s}{c}} = \frac{\tfrac{s}{c}}{\tfrac{s-c}{s}} + \frac{\tfrac{c}{s}}{\tfrac{c-s}{c}}$$
$$= \frac{s^2}{c(s-c)} + \frac{c^2}{s(c-s)} = \frac{s^2}{c(s-c)} - \frac{c^2}{s(s-c)}$$
$$= \frac{1}{s-c}\left(\frac{s^2}{c} - \frac{c^2}{s}\right) = \frac{1}{s-c}\cdot\frac{s^3 - c^3}{sc}$$
$$= \frac{(s-c)(s^2+sc+c^2)}{sc(s-c)} = \frac{s^2+c^2+sc}{sc}$$
$$= \frac{1 + sc}{sc} \quad (\because s^2+c^2=1)$$
$$= \frac{1}{sc} + 1 = 1 + \sec A\,\text{cosec}\,A = \textbf{RHS}$$
Hence proved. $\blacksquare$
Source: Exercise 8.3, Q4(iii), Chapter 8
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