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The expressions $\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}$ can be written as:

  1. $\sin A \ \cos A+1$
  2. $\sec A \ cosec A+1$
  3. $\tan A+ \cot A+1$
  4. $\sec A +cosec A$
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Ans is option (C)

Substitute  $cotA=\frac{1}{tanA}$  in the above question, we get

$\frac{tan^{2}A}{tanA-1}+\frac{1}{tanA(1-tanA)}$

Simplifying further we get,    $\frac{1}{tanA-1}[tan^{2}A-\frac{1}{tanA}]$

$\Rightarrow$   $\frac{tan^{3}A-1}{(tanA-1)tanA}$

$\Rightarrow$   $\frac{(tanA-1)(tan^{2}A+tanA+1)}{(tanA-1)tanA}$     (using  $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$)

$\Rightarrow$  $tanA+cotA+1$
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