The value of $$(\frac{1+sin\theta}{cos\theta}+\frac{cos\theta}{1+sin\theta})-2tan^{2}\theta$$ is

$$(\frac{1+sin\theta}{cos\theta}+\frac{cos\theta}{1+sin\theta})$$ - 2t$$an^2 \theta$$

$$\frac{(1+sin\theta)^2 + (cos\theta)^2}{cos\theta(1+sin\theta)}$$ - 2$$tan^{2}\theta$$

$$\frac{1+(sin\theta)^2 + (cos\theta)^2 + 2 sin\theta cos\theta}{cos\theta(1+sin\theta)}$$ - 2$$tan^{2}\theta$$

using $$sin^2 \theta + cos^2 \theta$$ = 1

$$\frac{1+1 + 2 sin\theta}{cos\theta(1+sin\theta)}$$ - 2$$tan^{2}\theta$$

$$\frac{2(1 + sin\theta)}{cos\theta(1+sin\theta)}$$ - 2$$tan^{2}\theta$$

2$$sec\theta$$ - 2$$tan^{2}\theta$$