Trigonometry
Root of equation via inverse trig; product of tangents
MMTS_Full_Test_19
Grade 12
Question:
One root of $\tan^{-1}\!\cot\!\left(\dfrac{3x^2+3|x|+1}{x^2+|x|+1}\right)=\dfrac{\pi}{2}-\csc\!\cdot\!\csc^{-1}\!\left(\dfrac{3|x|+2}{|x|+1}\right)$ is $2\sin\theta$, $\theta\in\left(0,\dfrac{\pi}{2}\right)$. Value of $\tan\dfrac{7\theta}{9}\cdot\tan\dfrac{10\theta}{9}\cdot\tan\dfrac{13\theta}{9}$ is
(A) 0
(B) $-1$
(C) 1
(D) 1/3
Step-by-Step Solution
Key Concept: Simplify equation to $x^2-|x|-1=0$. Solution $|x|=(1+\sqrt5)/2$. Root $=2\sin\theta=(1+\sqrt5)/2\Rightarrow\sin\theta=\sin(3\pi/10)\Rightarrow\theta=3\pi/10$.
Product $=1$.
Correct Answer: (C) 1