Multiple choice

If $\alpha$ and $\beta(\alpha > \beta)$ are 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

  1. $\dfrac{3\pi}{8}$
  2. $\dfrac{5\pi}{8}$
  3. $\dfrac{7\pi}{8}$
  4. $\dfrac{\pi}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is x^2 - sqrt(2)x + (sqrt(3)-sqrt(2)) = 0? No, sqrt(3-2sqrt(2)) = sqrt((sqrt(2)-1)^2) = sqrt(2)-1. So x^2 - sqrt(2)x + sqrt(2)-1 = 0. Roots are 1 and sqrt(2)-1. Since alpha > beta, alpha = 1, beta = sqrt(2)-1. cos^-1(1) = 0. tan^-1(1) = pi/4. tan^-1(sqrt(2)-1) = pi/8. Sum = 0 + pi/4 + pi/8 = 3pi/8.

Keep practicing — related questions