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 $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$)
$\Rightarrow$ $tanA+cotA+1$