Since $C=90^\circ$
Then $A+B=90$
Take, for instance $A=B=45$
So the origin equation becomes 1
I basically tried inputting the values in every option and arrived at $\frac {r_1}{R}$
Where r is the circumradius and $r_1$ is the radius of the excircle on side $A$.
How do I prove it properly?
Hint:
$$S=\sin C+\sin B-\sin A=2\sin\dfrac{B+C}2\cos\dfrac{B-C}2-2\sin\dfrac A2\cos\dfrac A2$$
As $\dfrac{B+C}2=\dfrac\pi2-\dfrac A2$
$$S=2\cos\dfrac A2\left(\cos\dfrac{B-C}2-\cos\dfrac{B+C}2\right)=4\cos\dfrac A2\sin\dfrac B2\sin\dfrac C2$$
Now use this
Here $C=90^\circ$