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 :
Here, \[\tan \alpha +\tan \beta =p\] ?..(i) \[\tan \alpha \,\tan \beta =q\] ?..(ii) Hence \[\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\]\[=\frac{p}{1-q}\] \[\Rightarrow \,\,{{\sin }^{2}}(\alpha +\beta )=\frac{1-\cos [2(\alpha +\beta )]}{2}\] \[=\frac{1}{2}\left\{ 1-\frac{1-{{\tan }^{2}}(\alpha +\beta )}{1+{{\tan }^{2}}(\alpha +\beta )} \right\}\]\[=\frac{1}{2}\left[ 1-\frac{1-{{\left( \frac{p}{1-q} \right)}^{2}}}{1+{{\left( \frac{p}{1-q} \right)}^{2}}} \right]\] \[=\frac{1}{2}\left[ \frac{{{(1-q)}^{2}}+{{p}^{2}}-{{(1-q)}^{2}}+{{p}^{2}}}{{{(1-q)}^{2}}+{{p}^{2}}} \right]\] \[=\frac{{{p}^{2}}}{{{p}^{2}}+{{(1-q)}^{2}}}\].
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