<s>[INST] 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
 [/INST] HINT:
$$\dfrac54=1+\sin x+\cos x+\sin x\cos x=1+\sqrt2\cos\left(\frac\pi4-x\right)+\dfrac{2\cos^2\left(\dfrac\pi4-x\right)-1}2$$
Write $\cos\left(\frac\pi4-x\right)=u$
$$4u^2+4\sqrt2u-3=0$$
We need $-1\le u\le1$
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