Question

If $\alpha$ and $\beta$ $(\alpha >\beta )$ are the roots of the equation $x^2-\sqrt{2}x+\sqrt{3-2\sqrt{2}}=0,$ then the value of $({\cos }^{-1}\alpha +{\tan }^{-1}\alpha +{\tan }^{-1}\beta )$ is equal to

• A. $\frac{3\pi }{8}$
• B. $\frac{5\pi }{8}$
• C. $\frac{7\pi }{8}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is A $\frac{3\pi }{8}$

$x^2-\sqrt{2}x+\sqrt{3-2\sqrt{2}}=0$
$\Rightarrow x^2-\sqrt{2}x+\sqrt{(\sqrt{2}-1{)}^2}=0$
$\Rightarrow x^2-1-\sqrt{2}(x-1)=0$
$\Rightarrow (x-1)(x+1-\sqrt{2})=0$
$\Rightarrow x=1,\sqrt{2}-1$
$\therefore \alpha =1$ and $\beta =\sqrt{2}-1$

Hence, ${\cos }^{-1}\alpha +{\tan }^{-1}\alpha +{\tan }^{-1}\beta =0+\frac{\pi }{4}+\frac{\pi }{8}=\frac{3\pi }{8}$