$$\left(\sec\phi-\tan\phi\right)^2\left(1+\sin\phi\right)^2\div sin^2 \phi$$ = ?

$$\left(\sec\phi-\tan\phi\right)^2\left(1+\sin\phi\right)^2\div sin^2 \phi$$
Put the $$\theta = 45\degree$$,
= $$\left(\sqrt{2}-1\right)^2\left(1+\frac{1}{\sqrt{2}}\right)^2\div \frac{1}{{2}}$$
= $$\left(\sqrt{2}-1\right)^2\left(\frac{\sqrt{2} + 1}{\sqrt{2}}\right)^2\div \frac{1}{{2}}$$
= $$\left(\sqrt{2}-1\right)^2\left(\sqrt{2} + 1\right)^2$$
= $$(2 + 1 - 2\sqrt{2})(2 + 1 + 2\sqrt{2})$$
= $$(3 - 2\sqrt{2})(3 + 2\sqrt{2})$$
= 9 - 8 = 1
From the option B,
$$\cot^2 \phi$$
Put the $$\theta = 45\degree$$,
= $$\cot^2 45\degree$$
= 1