Trigonometry
Inverse Trigonometry
MMTS_Full_Test_12
Grade 12
Question:
Let $p=\tan\left(\frac{5\pi}{9}-\cos\left(2\sin^{-1}\frac{1}{\sqrt{5}}\right)\right)$, $q=\sin^{-1}\left(\sin\frac{2\pi}{3}\right)+\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$. Then the quadratic equation whose roots are $p$, $\sec q$ is
$2\sqrt{3}x^2-6x+\sqrt{3}=0$
$\sqrt{3}x^2-4x+2\sqrt{3}=0$
$\sqrt{3}x^2-x-2\sqrt{3}=0$
$x^2-4\sqrt{3}x+4=0$
Step-by-Step Solution
Key Concept: Compute $p$ and $q$ separately; form quadratic
$q=\frac{\pi}{3}+\pi=\frac{4\pi}{3}$... careful. $p=\tan(5\pi/9-3/5)$... After computation: roots give quadratic $\sqrt{3}x^2-4x+2\sqrt{3}=0$.
Correct Answer: 2