Complex Numbers And Quadratic Equations question 598
Question: Given that $ \tan \alpha $ and $ \tan \beta $ are the roots of $ x^{2}-px+q=0, $ then the value of $ {{\sin }^{2}}(\alpha +\beta )= $ [RPET 2000]
Options:
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}}} $
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Answer:
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}{ 1-\frac{1-{{\tan }^{2}}(\alpha +\beta )}{1+{{\tan }^{2}}(\alpha +\beta )} } $ $ =\frac{1}{2}[ 1-\frac{1-{{( \frac{p}{1-q} )}^{2}}}{1+{{( \frac{p}{1-q} )}^{2}}} ] $ $ =\frac{1}{2}[ \frac{{{(1-q)}^{2}}+p^{2}-{{(1-q)}^{2}}+p^{2}}{{{(1-q)}^{2}}+p^{2}} ] $ $ =\frac{p^{2}}{p^{2}+{{(1-q)}^{2}}} $ .
BETA
