Let $$(1+\sin\theta)(1+\cos\theta) = \frac 54$$ Then what will be the value of $(1-\sin\theta)(1-\cos\theta)$.
I tried Squaring and Expanding the terms and Tried to replace the value with the expanded value from first equation but it got complicated further more.
We know $$\sin\theta=\frac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$$ $$\cos\theta=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$$ Set $x=\tan\frac{\theta}{2} $. We have $$\left(\frac{x^2+1}{x+1}\right)^2=1.6$$
$$(1+\sin x)(1+\cos x)=5/4\\\implies \sin x+\cos x=5/4-1-\sin x\cos x\\\implies 1+2\sin x\cos x=1/16+\sin ^2x\cos^2 x-(\sin x\cos x)/2$$
A quadratic in $\sin x\cos x$. Solve and you get values of $\sin x \cos x$
Square and replace $\cos^2$ with $(1-\sin^2),$ and you get value(s) of $\sin x$
$$(1+\sin x)(1+\cos x)=5/4 \Rightarrow 1+\sin x+\cos x+\frac{\sin 2x}{2} =5/4 \Rightarrow \sin x+\cos x+\frac{\sin 2x}{2} =1/4...(1)$$
Also,$$1-\sin x-\cos x+\frac{\sin 2x}{2} =y...(2)$$ And $$(\sin x+\cos x)^2 =1+\sin 2x...(3)$$
Substitute the term $\sin 2x$ from $(3)$ in $(1)$. Then we have $$\sin x+\cos x+\frac{(\sin x+\cos x)^2-1}{2} = \frac{1}{4}$$ Put $\sin x+\cos x=t$ and solve the resultant quadratic. We then have $t=-1\pm \frac{\sqrt{10}}{2}$. Since $t$ is positive, the admissible value of $t$ is $-1+\frac{\sqrt{10}}{2}$. Thus, $$y=1-t +\frac{t^2-1}{2}$$ from $(2)$. Hope it helps.
Consider the two equations
$$\begin{align} 1+s+c+sc&=G\\ 1-s-c+sc&=U \end{align}$$
where $G$ represents a Given number, $U$ represents an Unknown number, and we are using abbreviations $s=\sin\theta$ and $c=\cos\theta$. Summing the two equations gives $2(1+sc)=G+U$, from which we extract
$$2sc=G+U-2$$
Subtracting the two equations and then squaring gives
$$4(s^2+c^2+2sc)=G^2-2GU+U^2$$
If we now make the substitutions $s^2+c^2=1$ and $2sc=G+U-2$, we get
$$4(G+U-1)=G^2-2GU+U^2$$
or
$$U^2-(2G+4)U+G^2-4G+4=0$$
which solves to
$$U=G+2\pm\sqrt{8G}$$
For the problem at hand, $G=5/4$, which gives $U={13\over4}\pm\sqrt{10}$. However, $(1-\sin\theta)(1-\cos\theta)$ is clearly less than $4$, so the only meaningful value for $U$ is the one with the minus sign,
$$(1-\sin\theta)(1-\cos\theta)={13\over4}-\sqrt{10}$$
