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