If sin⁡θ>cos⁡θ, then the value of $$\frac{\sqrt{1-2\sin⁡ θ \cos ⁡θ}+\sqrt{1+2\sin ⁡θ\cos ⁡θ}}{2\tan⁡ θ}$$​ is:

Correct option is C

Given: 

​$$\frac{\sqrt{1-2\sin⁡ θ \cos ⁡θ}+\sqrt{1+2\sin ⁡θ\cos ⁡θ}}{2\tan⁡ θ}$$ 

Formula Used:​

​$$\sin^2 \theta + \cos^2 \theta = 1 \\ \ \\ (a+b)^2 = a^2+b^2+2ab \\ \ \\(a-b)^2 = a^2+b^2-2ab \\ \ \\ \tan \theta = \frac{\sin \theta}{\cos \theta}$$ 

Solution:  

$$\frac{\sqrt{1-2\sin⁡ θ \cos ⁡θ}+\sqrt{1+2\sin ⁡θ\cos ⁡θ}}{2\tan⁡ θ} \\ \ \\ = \frac{\sqrt{\sin^2\theta+\cos^2\theta-2\sin⁡ θ \cos ⁡θ}+\sqrt{\sin^2 \theta +\cos^2 \theta+2\sin ⁡θ\cos ⁡θ}}{2\tan⁡ θ} \\ \ \\ = \frac{\sqrt{(\sin\theta- \cos\theta)^2}+\sqrt{(\sin \theta +\cos \theta)^2}}{2\tan⁡ θ} \\ \ \\ = \frac{\sin\theta- \cos\theta+\sin \theta +\cos \theta}{2\tan⁡ θ}\\ \ \\ = \frac{2\sin\theta}{2\tan⁡ θ} \\ \ \\ = \cos \theta$$​​