A) \[L{{i}^{+}}\]
B) \[N{{a}^{+}}\]
C) \[{{K}^{+}}\]
D) \[{{C}_{2}}{{H}_{4}}B{{r}_{2}}\xrightarrow[{}]{Alc.KOH}{{C}_{2}}{{H}_{2}}\]
Correct Answer: A
Solution :
Since, tan A and tan B are the roots of the equation \[{{x}^{2}}-ax+b=0.\]Therefore, \[\tan A+\tan B=a\]and \[\tan A\tan B=b\] Now, \[\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{a}{1-b}\] Now,\[{{\sin }^{2}}(A+B)=\frac{1}{2}[1-\cos 2(A+B)]\] \[\Rightarrow \]\[{{\sin }^{2}}(A+B)=\frac{1}{2}\left\{ 1-\frac{1-{{\tan }^{2}}(A+B)}{1+{{\tan }^{2}}(A+B)} \right\}\] \[\Rightarrow \]\[{{\sin }^{2}}(A+B)=\frac{1}{2}\left\{ 1-\frac{1-\frac{{{a}^{2}}}{{{(1-b)}^{2}}}}{1+\frac{{{a}^{2}}}{{{(1-b)}^{2}}}} \right\}\] \[\Rightarrow \]\[{{\sin }^{2}}(A+B)=\frac{1}{2}\left\{ \frac{2{{a}^{2}}}{{{a}^{2}}{{(1-b)}^{2}}} \right\}\] \[\Rightarrow \]\[{{\sin }^{2}}(A+B)=\frac{{{a}^{2}}}{{{a}^{2}}+{{(1-b)}^{2}}}\]
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