Find the value of $$\frac{(\cot\theta-cosec\theta+1)(\tan\theta+\sec\theta+1)}{\cos\theta cosec\theta}?$$

Expression : $$\frac{(\cot\theta-cosec\theta+1)(\tan\theta+\sec\theta+1)}{\cos\theta cosec\theta}$$

= $$\frac{sin\theta}{cos\theta}\times[(\frac{cos\theta}{sin\theta}-\frac{1}{sin\theta}+1)\times(\frac{sin\theta}{cos\theta}+\frac{1}{cos\theta}+1)]$$

= $$\frac{sin\theta}{cos\theta}\times[(\frac{cos\theta+sin\theta-1}{sin\theta})\times(\frac{cos\theta+sin\theta+1}{cos\theta})]$$

Using, $$(x-y)(x+y)=x^2-y^2$$, where $$x=cos\theta+sin\theta$$ and $$y=1$$

= $$\frac{1}{cos^2\theta}\times[(cos\theta+sin\theta)^2-(1)^2]$$

= $$\frac{1}{cos^2\theta}\times[cos^2\theta+sin^2\theta+2cos\theta.sin\theta-1]$$

= $$\frac{1}{cos^2\theta}\times[1+2cos\theta.sin\theta-1]$$

= $$\frac{1}{cos^2\theta}\times(2cos\theta.sin\theta)$$

= $$\frac{2sin\theta}{cos\theta}=2tan\theta$$

=> Ans - (D)