Question:
If $3\sin^{-1}\!\left(\dfrac{2x}{1+x^{2}}\right) - 4\cos^{-1}\!\left(\dfrac{1-x^{2}}{1+x^{2}}\right) + 2\tan^{-1}\!\left(\dfrac{2x}{1-x^{2}}\right) = \dfrac{\pi}{3}$, then $x$ equals
If $3\sin^{-1}\!\left(\dfrac{2x}{1+x^{2}}\right) - 4\cos^{-1}\!\left(\dfrac{1-x^{2}}{1+x^{2}}\right) + 2\tan^{-1}\!\left(\dfrac{2x}{1-x^{2}}\right) = \dfrac{\pi}{3}$, then $x$ equals
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Key identities ($|x| \le 1$): $\sin^{-1}\tfrac{2x}{1+x^2} = 2\tan^{-1}x$; $\cos^{-1}\tfrac{1-x^2}{1+x^2} = 2\tan^{-1}x$; $\tan^{-1}\tfrac{2x}{1-x^2} = 2\tan^{-1}x$.
Updated On: May 2, 2026
- $\dfrac{1}{\sqrt{3}}$
- $-\dfrac{1}{\sqrt{3}}$
- $\sqrt{3}$
- $-\dfrac{\sqrt{3}}{2}$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Use the standard substitution identities for $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ in terms of $\tan^{-1}x$.
Step 2: Detailed Explanation:
For $|x| \le 1$: $\sin^{-1}\dfrac{2x}{1+x^2} = 2\tan^{-1}x$, $\;\cos^{-1}\dfrac{1-x^2}{1+x^2} = 2\tan^{-1}x$, $\;\tan^{-1}\dfrac{2x}{1-x^2} = 2\tan^{-1}x$.
LHS $= 3(2\tan^{-1}x) - 4(2\tan^{-1}x) + 2(2\tan^{-1}x) = (6-8+4)\tan^{-1}x = 2\tan^{-1}x$.
$2\tan^{-1}x = \dfrac{\pi}{3} \Rightarrow \tan^{-1}x = \dfrac{\pi}{6} \Rightarrow x = \dfrac{1}{\sqrt{3}}$.
Step 3: Final Answer:
$x = \dfrac{1}{\sqrt{3}}$.
Use the standard substitution identities for $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ in terms of $\tan^{-1}x$.
Step 2: Detailed Explanation:
For $|x| \le 1$: $\sin^{-1}\dfrac{2x}{1+x^2} = 2\tan^{-1}x$, $\;\cos^{-1}\dfrac{1-x^2}{1+x^2} = 2\tan^{-1}x$, $\;\tan^{-1}\dfrac{2x}{1-x^2} = 2\tan^{-1}x$.
LHS $= 3(2\tan^{-1}x) - 4(2\tan^{-1}x) + 2(2\tan^{-1}x) = (6-8+4)\tan^{-1}x = 2\tan^{-1}x$.
$2\tan^{-1}x = \dfrac{\pi}{3} \Rightarrow \tan^{-1}x = \dfrac{\pi}{6} \Rightarrow x = \dfrac{1}{\sqrt{3}}$.
Step 3: Final Answer:
$x = \dfrac{1}{\sqrt{3}}$.
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