A) \[\frac{{{p}^{2}}}{\{{{p}^{2}}+{{(1-q)}^{2}}\}}\]
B) \[\frac{{{q}^{2}}}{({{p}^{2}}+{{q}^{2}})}\]
C) \[\frac{{{q}^{2}}}{\{{{p}^{2}}-(1-{{q}^{2}})\}}\]
D) \[\frac{{{p}^{2}}}{({{p}^{2}}+{{q}^{2}})}\]
Correct Answer: A
Solution :
Since tan A and tan B are the roots of the equation \[{{x}^{2}}-px+q=0.\] Then \[\tan \,A+\tan \,B=P\] and \[\tan \,A\,tan\,B=q\] Now, \[\tan (A+B)=\frac{\tan \,A+\tan \,B}{1-\tan \,A\,\tan \,B}\] \[=\frac{P}{1-q}\] \[\therefore \] \[{{\sin }^{2}}(A+B)=\frac{1-\cos [2(A+B)]}{2}\] \[=\frac{1}{2}\left[ 1-\frac{1-{{\tan }^{2}}\,(A+B)}{1+{{\tan }^{2}}(A+B)} \right]\] \[=\frac{1}{2}\left[ 1-\frac{1-{{\left( \frac{P}{1-q} \right)}^{2}}}{1+{{\left( \frac{P}{1-q} \right)}^{2}}} \right]\] \[=\frac{1}{2}\left[ \frac{{{(1-q)}^{2}}+{{p}^{2}}-{{(1-q)}^{2}}+{{p}^{2}}}{{{(1-q)}^{2}}+{{p}^{2}}} \right]\] \[=\frac{{{p}^{2}}}{{{p}^{2}}+{{(1-q)}^{2}}}\]
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