Inverse Trigonometry
Solution set of cos⁻¹ = tan⁻¹ equation
MMTS_Full_Test_06
Grade 12
Question:
Let set $A$ denote the solutions of $\cos^{-1}(4x^3-3x)=\tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right)$. Then
(A) $n(A)=0$
(B) $n(A)=1$ and only one element $< -\frac{1}{2}$
(C) $n(A)=1$ and only one element $>\frac{1}{2}$
(D) $n(A)=2$ and absolute value of both elements $>\frac{1}{2}$
Step-by-Step Solution
Key Concept: Use $\cos^{-1}(4x^3-3x)=3\cos^{-1}x$ (for $x\in[-1/2,1]$) and $\tan^{-1}(2x/(1-x^2))=2\tan^{-1}x$ (for $|x|<1$). Equation becomes $3\cos^{-1}x=2\tan^{-1}x$.
$n(A)=1$, one element $>1/2$.
Correct Answer: (C) $n(A)=1$ and only one element $>\frac{1}{2}$