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) \[\sin (\alpha +\beta )=-p\]
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 \[={{\cos }^{2}}(\alpha +\beta )\,\,[{{\tan }^{2}}(\alpha +\beta )+p\tan \,(\alpha +\beta )+q]\] \[=\frac{1}{1+{{\tan }^{2}}(\alpha +\beta )}\,\left[ \frac{{{p}^{2}}}{{{(q-1)}^{2}}}+\frac{{{p}^{2}}}{q-1}+q \right]\] \[=\frac{{{(q-1)}^{2}}}{{{(q-1)}^{2}}+{{p}^{2}}}\left[ \frac{{{p}^{2}}+{{p}^{2}}(q-1)+q\,{{(q-1)}^{2}}}{{{(q-1)}^{2}}} \right]\] \[=\frac{q\,\left\{ {{p}^{2}}+{{(q-1)}^{2}} \right\}}{{{p}^{2}}+{{(q-1)}^{2}}}=q=R.H.S.\] of i.e., relation given in (a) is also satisfied.
You need to login to perform this action.
You will be redirected in
3 sec
