Question

If $sin\theta$ and $cos\theta$ are the roots of $a{x}^2+bx+c=0,$ then $co{s}^{-1}(a^2-b^2+2ac)=$

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. $0$

Answer 1

The correct option is B $\frac{\pi }{2}$

$cos\theta +sin\theta =-\frac{b}{a},cos\theta sin\theta =\frac{c}{a}\Rightarrow \frac{b^2}{a^2}=1+2\frac{c}{a}\Rightarrow b^2=a^2+2ac\Rightarrow a^2-b^2+2ac=0\Rightarrow co{s}^{-1}(a^2-b^2+2ac)=co{s}^{-1}0=\frac{\pi }{2}$