If $\left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,$ Then $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right)$ If $\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,$ Then $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) = $
$\bf{My\; Try::}$ Given $\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\Rightarrow 1+\sin \phi\cdot \cos \phi+\sin \phi+\cos \phi = \frac{5}{4}.$
So $\displaystyle \sin \phi+\cos \phi+\sin \phi \cdot \cos \phi = \frac{1}{4}\Rightarrow \left(\sin \phi+\cos \phi\right) = \frac{1}{4}-\sin \phi \cdot \cos \phi.$
Now $\displaystyle \left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) = 1-\left(\sin \phi+\cos \phi\right)+\sin \phi \cdot \cos \phi =\frac{3}{4}+\sin 2\phi$
Now How can I calculate $\sin 2\phi.$
Help me, Thanks

Like Sameer Kailasa,
$\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \dfrac{5}{4}\ \ \ \ (1)$ 
and let $\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) =x \ \ \ \ (2)$
$(1)-(2)\implies2(\sin \phi+\cos\phi)=\dfrac{5}{4}-x$ 
$\implies \sin \phi+\cos\phi=\dfrac{5-4x}8 $
Squaring we get $1+2\sin \phi\cos\phi=\left(\dfrac{5-4x}8\right)^2\iff2\sin \phi\cos\phi=\left(\dfrac{5-4x}8\right)^2-1$
$(1)+(2)\implies2(1+\sin \phi\cos\phi)=x+\dfrac{5}{4}\implies 2\sin \phi\cos\phi=\dfrac{4x-3}4$
Equate the values of $2\sin \phi\cos\phi$
