Will you solve for $\theta$ ?

Let $\left( 1- \sin \theta \right) \left( 1- \cos \theta \right) = x$ .

Now, upon expanding all terms we get,

$\left( 1+ \sin \theta \right) \left( 1+ \cos \theta \right) = 1+ \sin \theta + \cos \theta + \sin \theta \cos \theta = \dfrac{5}{4}$ $...\left( 1 \right)$

$\left( 1- \sin \theta \right) \left( 1- \cos \theta \right) = 1- \sin \theta - \cos \theta + \sin \theta \cos \theta = x$ $...\left( 2 \right)$

Adding $\left( 1 \right)$ and $\left( 2 \right)$ , we get,

$2 + 2\sin \theta \cos \theta = \dfrac{5}{4} + x$

$\large \therefore$ $\sin \theta \cos \theta = \dfrac{4x-3}{8}$ $...\left( 3 \right)$

Multiplying $\left( 1 \right)$ and $\left( 2 \right)$ , we get,

$\left( 1- \sin^{2} \theta \right) \left( 1- \cos^{2} \theta \right) = \dfrac{5}{4} x$

$\large \therefore$ $\sin^{2} \theta \cos^{2} \theta = \dfrac{5}{4} x$ $...\left( 4 \right)$

From $\left( 3 \right)$ and $\left( 4 \right)$ , we get,

${\left( \dfrac{4x-3}{8} \right)}^{2} = \dfrac{5}{4} x$

Solving for $x$ , we get,

$x = \dfrac{13 \pm 4\sqrt{10}}{4}$

From $\left( 4 \right)$ , we have,

$\sin^{2} \theta \cos^{2} \theta = \dfrac{5}{4} x$

$\Longrightarrow x = \dfrac{4}{5} \sin^{2} \theta \cos^{2} \theta$

$\therefore x$ can have a maximum value of $\dfrac{1}{5}$ . $\left( Why? \right)$

So, from the values of $x$ we got upon solving the quadratic equation only $x = \boxed {\dfrac{13 - 4\sqrt{10}}{4}}$ satisfies the criteria.

One way to solve it would be to solve for $\theta$ . I used a different technique and got a bit lucky.

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

$(cos^{2}\theta)(sin^{2}\theta)$ = $\frac{5}{4}(1-sin\theta)(1-cos\theta)$

$\frac{4}{5}(cos^{2}\theta)(sin^{2}\theta)$ = $(1-sin\theta)(1-cos\theta)$

Since $sin\theta$ and $cos\theta$ can be no larger than $1$ , LHS can be no larger than $\frac{4}{5}$ . As it turns out, there is only one option in the list that is smaller than $\frac{4}{5}$

Let (1-sinx)(1-cosx) = y

1-sinx-cosx+sinxcosx=y

We are given 1+sinx+cosx+sinxcosx = 5/4

Subtract these to get 5/4 - y = 2(sinx+cosx)

Now (1+sinx)(1+cosx) = 5/4

We know 1+cosx = 2cos^2(x/2)

1+sinx = (sinx/2 + cosx/2)^2

Substituting and simplifying

We get 5/2 = (sinx+cosx+1)^2

From here sinx+cosx can be found out and thats what we require