<s>[INST] 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 $\alpha + \beta=\frac {\pi}{2}$
 [/INST] 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  $\tan\dfrac{A+B}2=\tan^2\dfrac{A-B}2$ 
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