Trigonometry (Triangle Condition)

$\def\D{\displaystyle}$
Question (1): In $\Delta ABC$, if $\cot A+\cot B+\cot C=\sqrt 3$, then $\Delta ABC$ is equilateral.

$\D \begin{array}{|rl|}\hline \cot(A+B)&=\D\frac{\cot A\cot B-1}{\cot A+\cot B}\\ \hline \end{array}$

Proof:
$\cot C=\cot(180-(B+C))=-\cot(A+B)=\frac{1-\cot A\cot B}{\cot A+\cot B}.$
Let $\cot A=x,\cot B=y.$ Hence,
\begin{eqnarray*}
x+y+\frac{1-xy}{x+y}&=&\sqrt 3\\
(x^2+2xy+y^2) +(1-xy)&=&\sqrt 3x+\sqrt 3y\\
\end{eqnarray*}
For real  solutions, $b^2-4ac\ge 0.$ Thus
\begin{eqnarray*}
(x-\sqrt 3)^2-4(1)(x^2-\sqrt 3x+1)&\ge 0\\
3x^2-2\sqrt 3x+1&\le&0\\
(\sqrt 3x-1)^2&\le&0\\
\sqrt 3x-1&=&0.
\end{eqnarray*}
Thus $x=1/\sqrt 3$. By (1), $y=1/\sqrt 3$. Therefore
$\cot A=\cot B=\frac{1}{\sqrt 3}\Longrightarrow A=B=60^{\circ}.$

Question (2): In $\Delta ABC$, if $\sin^2 A+\sin^2 B+\sin^2 C= 2$, then $\Delta ABC$ is a right triangle.
$\D\begin{array}{|rl|}\hline \sin^2A&=\D \frac{1-\cos2A}{2}\\ \cos A+\cos B&=\D2\cos\frac{A+B}{2}\cos\frac{A-B}{2}\\ \sin^2A&=1-\cos^2A\\ \sin A&=\sin(180^{\circ}-A)\\ \hline \end{array}$

Proof:
\begin{eqnarray*}
\sin^2A+\sin^2B&=&\frac{1-\cos2A}{2}+\frac{1-\cos2B}{2} \\
&=&1-\frac{1}{2}(\cos2A+\cos2B)\\
&=&1-\frac{1}{2}\left(2\cos \frac{2A+2B}{2}\cos \frac{2A-2B}{2}\right)\\
\sin^2A+\sin^2B &=&1-\cos(A+B)\cos(A-B)\cdots (1)\\
\sin^2C&=&\sin^2(180^{\circ}-C)=\sin^2(A+B)\\
\sin^2C&=&1-\cos^2(A+B)\cdots (2)
\end{eqnarray*}
(1)+(2):
$\D\sin^2A+\sin^2B+\sin^2C$\begin{eqnarray*}
&=&2-\cos(A+B)(\cos(A+B)+\cos(A-B))\\
2&=&2-\cos(A+B)\left(\cos A\cos B-\sin A\sin B \right.\\
Hence, $\D \cos(A+B)=0$ or $\D \cos A=0$ or $\D\cos B=0.$
Thus  $\D A+B=90^{\circ}$, ie $\D C=90^{\circ}$ or $\D A=90^{\circ}$ or $\D B=90^{\circ}.$