What is the simplified value of $$\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}-\frac{2}{sin2A}$$ ?

Expression : $$\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}-\frac{2}{sin2A}$$

= $$\frac{tanA(1-tanA)+cotA(1-cotA)}{(1-tanA)(1-cotA)}-\frac{2}{2sinAcosA}$$

= $$\frac{tanA+cotA-(tan^2A+cot^2A)}{1-tanA-cotA+tanAcotA}-\frac{1}{sinAcosA}$$      

= $$\frac{tanA+cotA-[(tanA+cotA)^2-2]}{1-(tanA+cotA)+1}-\frac{1}{sinAcosA}$$             [$$\because tanx.cotx=1$$]

Let $$(tanA+cotA)=x$$ -----------(i)

= $$\frac{x-x^2+2}{2-x}-\frac{1}{sinAcosA}$$

= $$\frac{(2-x)(x+1)}{2-x}-\frac{1}{sinAcosA}$$

= $$x+1-\frac{1}{sinAcosA}$$ ------------(ii)

From equation (i), => $$\frac{sinA}{cosA}+\frac{cosA}{sinA}=x$$

=> $$\frac{sin^2A+cos^2A}{sinAcosA}=x$$

=> $$\frac{1}{sinAcosA}=x$$

Substituting above value in equation (ii), we get :

= $$x+1-x=1$$

=> Ans - (C)