Trigonometric Identities Question 11
Question: If $ \tan \alpha ,\tan \beta $ are the roots of the equation $ x^{2}+px+q=0\text{ }(p\ne 0), $ then
Options:
A) $ {{\sin }^{2}}(\alpha +\beta )+p\sin (\alpha +\beta )\cos (\alpha +\beta )+q{{\cos }^{2}}(\alpha +\beta )=q $
B) $ \tan (\alpha +\beta )=\frac{p}{q-1} $
C) $ \cos (\alpha +\beta )=1-q $
D) $ \sin (\alpha +\beta )=-p $
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Answer:
Correct Answer: B
Solution:
Since $ \tan ,\alpha ,,\tan \beta $ are the roots of the equation $ x^{2}+px+q=0. $ \ $ \tan \alpha +\tan \beta =-p, $ $ \tan \alpha \tan \beta =q $ \ $ \tan ,(\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{p}{q-1} $ , which is given in . Also when $ \tan ,(\alpha +\beta )=\frac{p}{q-1} $ . L.H.S. of the expression given in $ ={{\cos }^{2}}(\alpha +\beta )[{{\tan }^{2}}(\alpha +\beta )+p\tan ,(\alpha +\beta )+q] $ $ =\frac{1}{1+{{\tan }^{2}}(\alpha +\beta )},[ \frac{p^{2}}{{{(q-1)}^{2}}}+\frac{p^{2}}{q-1}+q ] $ $ =\frac{{{(q-1)}^{2}}}{{{(q-1)}^{2}}+p^{2}}[ \frac{p^{2}+p^{2}(q-1)+q,{{(q-1)}^{2}}}{{{(q-1)}^{2}}} ] $ $ =\frac{q,{ p^{2}+{{(q-1)}^{2}} }}{p^{2}+{{(q-1)}^{2}}}=q=R.H.S. $ of i.e., relation given in (a) is also satisfied.
BETA
