Question

If $tan\alpha ,tan\beta$ are the roots of the equation $x^2$ + ax + b = 0 then the value of $si{n}^2(\alpha +\beta )+asin(\alpha +\beta )cos(\alpha +\beta )+bco{s}^2(\alpha +\beta )=$

• A. a
• B. b
• C. ab
• D. $\frac{1}{a}+b$

Answer 1

The correct option is B

b

$tan\alpha +tan\beta =-a,tan\alpha tan\beta =btan(\alpha +\beta )=\frac{tan\alpha +tan\beta }{1-tan\alpha tan\beta }tan(\alpha +\beta )=\frac{-a}{1-b}=\frac{a}{b-1}$
$Nowsi{n}^2(\alpha +\beta )+asin(\alpha +\beta )cos(\alpha +\beta )+bco{s}^2(\alpha +\beta )=co{s}^2(\alpha +\beta )\left[ta{n}^2\right(\alpha +\beta )+atan(\alpha +\beta )+b]=\frac{ta{n}^2(\alpha +\beta )+atan(\alpha +\beta )+b}{se{c}^2(\alpha +\beta )}=\frac{ta{n}^2(\alpha +\beta )+atan(\alpha +\beta )+b}{1+ta{n}^2(\alpha +\beta )}$
$=\frac{\frac{a^2}{(b-1{)}^2}+a.\frac{a}{b-1}+b}{1+\frac{a^2}{(b-1{)}^2}}=\frac{\frac{a}{b-1}(\frac{a}{b-1}+a)+b}{1+\frac{a^2}{(b-1{)}^2}}=\frac{\frac{a}{b-1}\left(\frac{a+ab-a}{b-1}\right)+b}{1+\frac{a^2}{(b-1{)}^2}}=b$