$a,b,c$ are the sides of a right triangle whose smallest angle is $\theta$ and $\frac 1a,\frac 1b,\frac 1c$ are the sides of another right triangle then what is the value of $\sin\theta$.
I got two relations
$$a^2+b^2=c^2$$ and $$c^2=\frac{a^2b^2}{a^2+b^2}$$ Which can be rewritten as
$(a^2+b^2)^2=a^2b^2$ or, $a^4+b^4+a^2b^2=0$
Now if $b= c \cos\theta$ & $a=c \sin\theta$:
Then,
$c^4\sin^4\theta+c^4\cos^4\theta+c^4\sin^2\theta\cos^2\theta=0$
$c^4{\sin^2\theta+\cos^2\theta}^2=2\sin^2\theta\cos^2\theta$
$c^4\cdot 1= 2\sin^2\theta\cos^2\theta$ I am lost after this.
Without loss of generality, assume $a\le b < c$.
Without affecting the conditions, we can scale $a,b,c$, so that $c=1$.
Then $a \le b < 1$, and $\sin(\theta)=a$.
Thus, our goal is to solve for $a$.
By the Pythagorean theorem, we get the equations \begin{align*} & \!\! \begin{cases} {\displaystyle{a^2+b^2=1}}\\[4pt] {\displaystyle{1+\frac{1}{b^2}=\frac{1}{a^2}}}\\ \end{cases} \\[10pt] \text{Then}\;\;&1+\frac{1}{b^2}=\frac{1}{a^2}\\[4pt] \implies\;&a^2b^2+a^2=b^2\\[4pt] \implies\;&a^2(1-a^2)+a^2=1-a^2&&\text{[since $b^2=1-a^2$]}\\[4pt] \implies\;&2a^2-a^4=1-a^2\\[4pt] \implies\;&-2a^2+a^4=a^2-1\\[4pt] \implies\;&1-2a^2+a^4=a^2\\[4pt] \implies\;&(1-a^2)^2=a^2\\[4pt] \implies\;&1-a^2=\pm{a}\\[4pt] \implies\;&1-a^2=a&&\text{[since $0 < a < 1$]}\\[4pt] \implies\;&a=\frac{-1\pm\sqrt{5}}{2}\\[4pt] \implies\;&a=\frac{\sqrt{5}-1}{2}&&\text{[since $a > 0$]}\\[4pt]\\[4pt] \end{align*}