Question

$\tan {22}^{\circ }$ and $\tan {23}^{\circ }$ are roots of $x^2+ax+b=0$ then

• A. $a+b+1=0$
• B. $a-b+1=0$
• C. $b-a+1=0$
• D. $a+b=1$

Answer 1

Answer: (B)

$a-b+1=0$

Let $\alpha =\tan {22}^o$ and $\beta =\tan {23}^o$

Where $\alpha$ and $\beta$ are roots of the equation $x^2+ax+b=0$

$\therefore$ $\alpha +\beta =\tan {22}^o+\tan {23}^o=-a$ ----- ( 1 )

$\Rightarrow$ $\alpha \beta =\tan {22}^o\times \tan {23}^o=b$ ---- ( 2 )

We know

$\tan {45}^o=\tan ({22}^o+{23}^o)=\frac{\tan {22}^o+\tan {23}^o}{1-\tan {22}^o.\tan {23}^o}$

From ( 1 ) and ( 2 ),

$\Rightarrow$ $\tan {45}^o=\frac{-a}{1-b}$

$\Rightarrow$ $1=\frac{-a}{1-b}$

$\Rightarrow$ $1-b=-a$

$\therefore$ $a-b+1=0$