Prove that $\measuredangle\gamma= 90^{\circ}$

Given a triangle, if $\sin^{2}\alpha+ \sin^{2}\beta=\!\sin\gamma, \max\!\left ( \measuredangle\alpha- l\measuredangle\beta, \measuredangle\beta- l\measuredangle\alpha \right )\leq 90^{\circ}, \left | l \right |\leq 3$ so $$\measuredangle\gamma= 90^{\circ}$$ with three angles $\alpha, \beta, \gamma$ and $l= constant.$

I have posted that triangle equality for so long.. also had a case proof$,\quad l= 0$ is the only successful one. Now let me show. By substitution$,\quad\sin\alpha:= \frac{2a\left ( a+ 1 \right )}{2a\left ( a+ 1 \right )+ 1}, \sin\beta:= \frac{2b+ 1}{2b^{2}+ 2b+ 1}$ with positives $a, b$ $$\Rightarrow\sin^{2}\alpha+ \sin^{2}\beta- \left ( \sin\gamma \right )_{= 1}= \frac{4\left ( a- b \right )\left ( a+ b+ 1 \right )\left ( 2ab+ a+ b \right )\left ( 2ab+ a+ b+ 1 \right )}{\left ( 2a^{2}+ 2a+ 1 \right )^{2}\left ( 2b^{2}+ 2b+ 1 \right )^{2}}\Rightarrow$$ $$\Rightarrow a= b\Rightarrow\sin^{2}\alpha+ \sin^{2}\beta= 1\Rightarrow\measuredangle\gamma= 90^{\circ}$$

This answer proves that

if there is a triangle with angles $\alpha,\beta$ and $\gamma$ such that $\sin^2\alpha+\sin^2\beta=\sin\gamma$ and $\max(\alpha−l\beta, \beta−l\alpha)\le\frac{\pi}{2}$ for some $l$ with $|l|\le 3$, then $\gamma=\frac{\pi}{2}$.

Proof :

Since $\gamma=\pi-\alpha-\beta$, we see that $\sin^{2}\alpha+ \sin^{2}\beta=\sin\gamma$ is equivalent to $$\sin^{2}\alpha+ \sin^{2}\beta=\sin(\alpha+\beta)\tag1$$

Here, let $k:=\alpha+\beta$. Since $\beta=k-\alpha$, we see that $(1)$ is equivalent to $$\begin{align}&\sin^2\alpha+\sin^2(k-\alpha)=\sin k \\\\&\iff \sin^2\alpha+(\sin k\cos\alpha-\cos k\sin\alpha)^2=\sin k \\\\&\iff \sin^2\alpha+\sin^2k\cos^2\alpha-2\sin k\cos\alpha\cos k\sin\alpha+\cos^2k\sin^2\alpha=\sin k \\\\&\iff \sin^2\alpha+\sin^2k(1-\sin^2\alpha)-2\sin k\cos\alpha\cos k\sin\alpha+\cos^2k\sin^2\alpha=\sin k \\\\&\iff (1-\sin^2k+\cos^2k)\sin^2\alpha-2\sin k\cos\alpha\cos k\sin\alpha=\sin k-\sin^2k \\\\&\iff 2\cos^2k\sin^2\alpha-2\sin k\cos\alpha\cos k\sin\alpha=\sin k-\sin^2k \\\\&\iff 2\cos k\sin\alpha\bigg(\cos k\sin\alpha-\sin k\cos\alpha\bigg)=\sin k-\sin^2k \\\\&\iff 2\cos k\sin\alpha\sin(\alpha-k)=\sin k-\sin^2k \\\\&\iff 2\cos k\bigg(-\frac 12\bigg)\bigg(\cos(2\alpha-k)-\cos k\bigg)=\sin k-\sin^2k \\\\&\iff \cos(2\alpha-k)\cos k=1-\sin k \end{align}$$

Here, let $m:=2\alpha-k$.

Then, since $\alpha,\beta,\gamma$ are written as $$\alpha=\frac{k+m}{2},\qquad\beta=\frac{k-m}{2},\qquad\gamma=\pi-k$$ our conditions

$$\begin{cases}\alpha\gt 0,\quad \beta\gt 0,\quad \gamma\gt 0 \\\alpha+\beta+\gamma=\pi \\\alpha- l\beta\le \frac{\pi}{2} \\\beta- l\alpha\leq \frac{\pi}{2} \\\left | l \right |\leq 3 \\\sin^2\alpha+\sin^2\beta=\sin\gamma\end{cases}$$

can be written as

$$\begin{cases}0\lt k\lt \pi \\-k\lt m\lt k \\ -(kl-k+\pi)\le (l+1)m\le kl-k+\pi \\\left | l \right |\leq 3 \\\cos m\cos k=1-\sin k\end{cases}$$

So, our problem is reduced to proving that

if there is $(k,l,m)$ such that$$\begin{cases}0\lt k\lt \pi \\-k\lt m\lt k \\ -(kl-k+\pi)\le (l+1)m\le kl-k+\pi \\\left | l \right |\leq 3 \\\cos m\cos k=1-\sin k\end{cases}$$ then $k=\frac{\pi}{2}$.

Case 1 : $-3\le l\lt -1$

$$\begin{cases}0\lt k\lt \pi \\-k\lt m\lt k \\ \frac{kl-k+\pi}{l+1}\le m\le \frac{-(kl-k+\pi)}{l+1} \\\cos m\cos k=1-\sin k\end{cases}$$

• Suppose that $0\lt k\lt\frac{\pi}{2}$. Then, we have $$\begin{align}\cos k\lt \cos m&\implies\cos k\lt \frac{1-\sin k}{\cos k} \\\\&\implies \cos^2k\lt 1-\sin k \\\\&\implies 1-\sin^2k\lt 1-\sin k \\\\&\implies \sin k(\sin k-1)\gt 0 \\\\&\implies \sin k\lt 0\end{align}$$ which contradicts that $0\lt k\lt\frac{\pi}{2}$.

• Suppose that $k=\frac{\pi}{2}$. Then, there is no $m$ such that $\frac{\pi}{2}=\frac{kl-k+\pi}{l+1}\le m\le \frac{-(kl-k+\pi)}{l+1}=-\frac{\pi}{2}$.

• Suppose that $\frac{\pi}{2}\lt k\lt \pi$. Then, $0\le\frac{-(kl-k+\pi)}{l+1}\lt\frac{\pi}{2}$ implies $\cos\bigg(\frac{-(kl-k+\pi)}{l+1}\bigg)\le \frac{1-\sin k}{\cos k}$ which is impossible since LHS is positive while RHS is negative.

So, we see that there is no triangle with angles $\alpha,\beta$ and $\gamma$ such that $\sin^2\alpha+\sin^2\beta=\sin\gamma$ and $\max(\alpha−l\beta, \beta−l\alpha)\le\frac{\pi}{2}$ for some $l$ with $-3\le l\lt -1$.

Case 2 : $l=-1$

$$\begin{cases}0\lt k\le \frac{\pi}{2} \\-k\lt m\lt k \\\cos m\cos k=1-\sin k\end{cases}$$

• $k=\frac{\pi}{2}$ is sufficient.

• Suppose that $0\lt k\lt \frac{\pi}{2}$. Then, we have $\cos k\lt \cos m$ which is impossible as seen in Case 1.

Case 3 : $-1\lt l\le 1$

$$\begin{cases}0\lt k\lt \pi \\-k\lt m\lt k \\ -\frac{kl-k+\pi}{l+1}\le m\le \frac{kl-k+\pi}{l+1} \\\cos m\cos k=1-\sin k\end{cases}$$

• $k=\frac{\pi}{2}$ is sufficient.

• Suppose that $0\lt k\lt \frac{\pi}{2}$. Then, we have $\cos k\lt \cos m$ which is impossible as seen in Case 1.

• Suppose that $\frac{\pi}{2}\lt k\lt\pi$. Then $0\le\frac{kl-k+\pi}{l+1}\le \frac{\pi}{2}\implies \cos\bigg( \frac{kl-k+\pi}{l+1}\bigg)\le \frac{1-\sin k}{\cos k}$ which is impossible since RHS is negative while LHS is non-negative.

Case 4 : $1\lt l\le 3$

$$\begin{cases}0\lt k\lt \pi \\-k\lt m\lt k \\ -\frac{kl-k+\pi}{l+1}\le m\le \frac{kl-k+\pi}{l+1} \\\cos m\cos k=1-\sin k\end{cases}$$

• $k=\frac{\pi}{2}$ is sufficient.

• Suppose that $0\lt k\lt \frac{\pi}{2}$. Then, we have $\cos k\lt \cos m$ which is impossible as seen in Case 1.

• Suppose that $\frac{\pi}{2}\lt k\lt\pi$. Then, $k\gt \frac{kl-k+\pi}{l+1}\ge 0$ implies $$\begin{align}&\cos\bigg( \frac{kl-k+\pi}{l+1}\bigg)\le \frac{1-\sin k}{\cos k} \\\\&\implies \cos\bigg( \frac{3k-k+\pi}{3+1}\bigg)\le\cos\bigg( \frac{kl-k+\pi}{l+1}\bigg)\le \frac{1-\sin k}{\cos k} \\\\&\implies \cos\bigg(\frac k2+\frac{\pi}{4}\bigg)\le \frac{1-\sin k}{\cos k} \\\\&\implies \frac{1}{\sqrt 2}\cos\frac k2-\frac{1}{\sqrt 2}\sin\frac k2\le\frac{1-\sin k}{\cos k} \\\\&\implies (\sqrt{1-s^2}-s)(1-2s^2)\ge \sqrt 2(1-2s\sqrt{1-s^2}) \\\\&\implies (1-2s^2+2\sqrt 2\ s)\sqrt{1-s^2}+2s^3-s-\sqrt 2\ge 0 \end{align}$$ where $s:=\sin\frac k2$ with $\frac{1}{\sqrt 2}\lt s\lt 1$. Now, let $$f(s):=(1-2s^2+2\sqrt 2\ s)\sqrt{1-s^2}+2s^3-s-\sqrt 2$$ In the following, let us prove $f(s)\lt 0$ for $\frac{1}{\sqrt 2}\lt s\lt 1$. Since we have $$f'(s)=\frac{(6 s^2 - 1)\sqrt{1 - s^2}+6 s^3 - 4\sqrt 2\ s^2 - 5 s + 2\sqrt 2}{\sqrt{1-s^2}}$$There is no $s$ such that $f'(s)=0$ and $\frac{1}{\sqrt 2}\lt s\lt 1$ since $$\begin{align}f'(s)=0&\implies 6 s^3 - 4\sqrt 2 s^2 - 5 s +2\sqrt 2=\sqrt{1 - s^2}\ (1-6 s^2) \\\\&\implies (6 s^3 - 4\sqrt 2 s^2 - 5 s + 2\sqrt 2)^2=(1-s^2)(1-6s^2)^2 \\\\&\implies 72 s^6 - 48\sqrt 2\ s^5 - 76 s^4 + 64\sqrt 2\ s^3 + 6 s^2 - 20\sqrt 2\ s + 7=0 \\\\&\implies \bigg(s-\frac{1}{\sqrt 2}\bigg)^3 (2 s + \sqrt 2) (6 s + \sqrt 2 - 4) (6 s + \sqrt 2 + 4)=0 \\\\&\implies s=\pm\frac{1}{\sqrt 2},\frac{-\sqrt 2\pm 4}{6}\end{align}$$none of which satisfies $\frac{1}{\sqrt 2}\lt s\lt 1$. Since we have $f'(s)\lt 0$ with $f(1/\sqrt 2)=0$, we finally get $f(s)\lt 0$.
  
  Therefore, having $f(s)\ge 0$ is a contradiction.