​​prove that a^2+b^2+2ab=sin^2A is possible inly when a=b

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$$\begin{gathered} \mathrm{Your} \mathrm{question} \mathrm{seems} \mathrm{incomplete}. \mathrm{We} \mathrm{have} \\ {\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}={\sin }^2\mathrm{A} \\ \Rightarrow {\left(\mathrm{a}+\mathrm{b}\right)}^2={\sin }^2\mathrm{A} \\ \mathrm{Now} \mathrm{above} \mathrm{relation} \mathrm{is} \mathrm{possible} \mathrm{even} \mathrm{if} \mathrm{a}\ne \mathrm{b} \\ \mathrm{Suppose} \mathrm{we} \mathrm{take} \mathrm{a}=1 \mathrm{and} \mathrm{b}=1 \\ \mathrm{Hence} \\ \mathrm{we} \mathrm{get} \\ 1={\sin }^2\mathrm{A} \mathrm{which} \mathrm{is} \mathrm{possible}. \\ \mathrm{Hence} {\left(\mathrm{a}+\mathrm{b}\right)}^2={\sin }^2\mathrm{A} \mathrm{can} \mathrm{be} \mathrm{true} \mathrm{even} \mathrm{if} \mathrm{a}\ne \mathrm{b} \end{gathered}$$

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