Prove that $\alpha + \beta=\frac {\pi}{2}$ It is given that-

(1) $0<\alpha,\beta<90$.
(2) $\sin^2\alpha+\sin^ 2\beta=\sin(\alpha+\beta).$

Prove that $\alpha + \beta=\frac {\pi}{2}$
 A: NOTE: it's answer to initial question (without squares), downvoters.
It's very strightforward.
$$
\sin\alpha + \sin\beta = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\\
\sin(\alpha + \beta) = 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha+\beta}{2}\\
\sin\alpha + \sin\beta = \sin(\alpha + \beta)\Longrightarrow \sin\frac{\alpha+\beta}{2} = 0\text{ or }\cos\frac{\alpha-\beta}{2} = \cos\frac{\alpha+\beta}{2}
$$
In the first case
$$
\sin\frac{\alpha+\beta}{2} = 0\Longrightarrow \alpha + \beta = \pi n,0 < \alpha + \beta < \pi;
$$
there ares no solutions. If $\alpha,\beta \color{red}\le \pi/2$, then $\alpha + \beta = \pi$.
In the second case,
$$\cos\frac{\alpha-\beta}{2} = \cos\frac{\alpha+\beta}{2} \Longrightarrow 2\sin\frac\alpha2\sin\frac\beta2 = 0,
$$
and $\alpha=2\pi k$ or $\beta=2\pi m$, $k,m\in\mathbb Z$.
Anyway, your statement is false.
A: We have -

$\sin^2\alpha+\sin^2\beta=\sin(\alpha+\beta)$

$\implies \sin\alpha\cos\beta+cos\alpha\sin\beta=\sin^2\alpha+\sin^2\beta$
$\implies\displaystyle\sin\alpha(\cos\beta-\sin\alpha)+\sin\beta(\cos\alpha-\sin\beta)=0$
$\implies2 \sin\alpha\cdot \sin\left({\frac{\beta+\frac \pi2-\alpha}{2}}\right)\cdot \sin\left({\frac {\frac{\pi}{2}-\alpha-\beta}{2}}\right)+2\sin\beta\cdot \sin\left({\frac{\alpha+\frac{\pi}{2}-\beta}{2}}\right)\cdot \sin\left({\frac{\frac{\pi}{2}-\beta-\alpha}{2}}\right)=0$
$\implies\sin\alpha\cdot \sin\left(\frac{\pi}{4}+\frac{\beta-\alpha}{2}\right)\cdot \sin\left(\frac{\pi}{4}-\frac{\alpha+\beta}{2}\right)+\sin\beta\cdot \sin\left(\frac{\pi}{4}+\frac{\alpha-\beta}{2}\right)\cdot \sin\left(\frac{\pi}{4}-\frac{\alpha+\beta}{2}\right)=0$
$\implies\sin\left(\frac{\pi}{4}-\frac{\beta+\alpha}{2}\right)=0$
$\implies\frac{\pi}{4}-\frac{\alpha+\beta}{2}=0$
$\implies \alpha+\beta=\frac{\pi}{2}$
A: Equation $(2)$ says that
$$\sin\alpha(\sin\alpha-\cos\beta)=\sin\beta(\cos\alpha-\sin\beta)\ .$$
With $\beta:={\pi\over2}-\beta'$ we therefore have
$$\sin\alpha(\sin\alpha-\sin\beta')=\sin\beta(\cos\alpha-\cos\beta')\ .$$
When $\alpha\ne\beta'$ the two sides of the last equation have different signs.
A: Its given (by you) that ${sin}^2\alpha+{sin}^2\beta={sin}(\alpha+\beta) ..........eqn (1)$
But ${sin}^2\beta=1-{cos}^2\beta$
Slotting the identity into $eqn(1)$ gives: $${sin}^2\alpha+(1-{cos}^2\beta)={sin}(\alpha+\beta) $$
$$1-({cos}^2\beta-{sin}^2\alpha)={sin}(\alpha+\beta)$$
$$1-{cos}(\alpha+\beta)={sin}(\alpha+\beta)$$
$${sin}(\alpha+\beta)+{cos}(\alpha+\beta)=1$$
Let $(\alpha+\beta)=A$ Therefore $sinA+cosA=1$
Solving this further by squaring both sides gives:
$${sin}^2A+{cos}^2A+2sinAcosA=1$$
$$2sinAcosA=0$$
Therefore $sinA=0$ or $cos A=0$
This gives $A=0$ or $A=\frac{\pi}{2}$
Recall that $(\alpha+\beta)=A$
Therefore $\alpha+\beta=0$ or $\alpha+\beta=\frac{\pi}{2}$
A: Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\sin^2A+\sin^2B=1-(\cos^2A-\sin^2B)=1-\cos(A+B)\cos(A-B)$$
$$\implies\sin(A+B)+\cos(A+B)\cos(A-B)=1$$
Using Weierstrass substitution for $\sin(A+B),\cos(A+B)$ and setting $\tan\dfrac{A+B}2=t,$
$$[1+\cos(A-B)]t^2-2t+1-\cos(A-B)=0$$
Using Weierstrass substitution for $\cos(A-B),$
$$t=1,\tan^2\dfrac{A-B}2$$
If $\tan\dfrac{A+B}2=1,\dfrac{A+B}2=n\pi+\dfrac\pi2\iff A+B=?$ where $n$
But I'm not sure about  $\tan\dfrac{A+B}2=\tan^2\dfrac{A-B}2$ 
