The value of $$\frac{ sec\phi\left(1-\sin\phi\right)\left(\sin \phi +\cos \phi \right)\left(\sec \phi +\tan \phi \right)}{ \sin\phi\left(1+\tan\phi\right)+\cos\phi\left(1+\cot\phi\right)}$$ is equal to:

$$\frac{ sec\phi\left(1-\sin\phi\right)\left(\sin \phi +\cos \phi \right)\left(\sec \phi +\tan \phi \right)}{ \sin\phi\left(1+\tan\phi\right)+\cos\phi\left(1+\cot\phi\right)}$$
Put the value of $$\phi = 60\degree$$,
$$\frac{ sec60\degree\left(1-\sin60\degree\right)\left(\sin60\degree +\cos 60\degree \right)\left(\sec 60\degree +\tan 60\degree \right)}{ \sin60\degree\left(1+\tan60\degree\right)+\cos60\degree\left(1+\cot60\degree\right)}$$
= $$\frac{ 2\left(1-\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2} +\frac{1}{2} \right)\left(2 +\sqrt{3} \right)}{ \frac{\sqrt{3}}{2}\left(1+\sqrt{3}\right)+\frac{1}{2}\left(1+\frac{1}{\sqrt{3}}\right)}$$
= $$\frac{\left(2-\sqrt{3}\right)\left(\frac{\sqrt{3} + 1}{2} \right)\left(2 +\sqrt{3} \right)}{ \frac{\sqrt{3}}{2}\left(1 + \sqrt{3}\right)+\frac{1}{2}\left(\frac{1 + \sqrt{3}}{\sqrt{3}}\right)}$$
= $$\frac{\left(2-\sqrt{3}\right)\left({\sqrt{3} + 1} \right)\left(2 +\sqrt{3} \right)}{ \sqrt{3}\left(1 + \sqrt{3}\right)+\left(\frac{1 + \sqrt{3}}{\sqrt{3}}\right)}$$

= $$\frac{\sqrt{3} + 1}{ \sqrt{3}\left(1 + \sqrt{3}\right)+\left(\frac{1 + \sqrt{3}}{\sqrt{3}}\right)}$$

= $$\frac{1}{ \sqrt{3}+\frac{1}{\sqrt{3}}}$$ = $$\frac{\sqrt{3}}{4}$$
From the option D,
$$sin\phi cos\phi$$
Put the value of $$\phi = 60\degree$$,
= $$sin60\degree cos60\degree$$
= $$\frac{\sqrt{3}}{2} \times \frac{1}{2}$$

= $$\frac{\sqrt{3}}{4}$$