Q69 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics
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 $x = ?$
- A$\sqrt{3}$
- B$1$
- Correct answerC$\dfrac{1}{\sqrt{3}}$
- D$-1$
Solution
Assuming $0 < x < 1$, the standard inverse trigonometric identities are:
$\sin^{-1}\left(\dfrac{2x}{1 + x^2}\right) = 2\tan^{-1}x$
$\cos^{-1}\left(\dfrac{1 - x^2}{1 + x^2}\right) = 2\tan^{-1}x$
$\tan^{-1}\left(\dfrac{2x}{1 - x^2}\right) = 2\tan^{-1}x$
Substituting these into the given equation yields:
$3(2\tan^{-1}x) - 4(2\tan^{-1}x) + 2(2\tan^{-1}x) = \dfrac{\pi}{3}$
$6\tan^{-1}x - 8\tan^{-1}x + 4\tan^{-1}x = \dfrac{\pi}{3}$
$2\tan^{-1}x = \dfrac{\pi}{3}$
$\tan^{-1}x = \dfrac{\pi}{6}$
$x = \tan\left(\dfrac{\pi}{6}\right) = \dfrac{1}{\sqrt{3}}$
Since $x = \dfrac{1}{\sqrt{3}}$ satisfies the condition $0 < x < 1$, it is the valid solution.
Answer: $\dfrac{1}{\sqrt{3}}$
$\sin^{-1}\left(\dfrac{2x}{1 + x^2}\right) = 2\tan^{-1}x$
$\cos^{-1}\left(\dfrac{1 - x^2}{1 + x^2}\right) = 2\tan^{-1}x$
$\tan^{-1}\left(\dfrac{2x}{1 - x^2}\right) = 2\tan^{-1}x$
Substituting these into the given equation yields:
$3(2\tan^{-1}x) - 4(2\tan^{-1}x) + 2(2\tan^{-1}x) = \dfrac{\pi}{3}$
$6\tan^{-1}x - 8\tan^{-1}x + 4\tan^{-1}x = \dfrac{\pi}{3}$
$2\tan^{-1}x = \dfrac{\pi}{3}$
$\tan^{-1}x = \dfrac{\pi}{6}$
$x = \tan\left(\dfrac{\pi}{6}\right) = \dfrac{1}{\sqrt{3}}$
Since $x = \dfrac{1}{\sqrt{3}}$ satisfies the condition $0 < x < 1$, it is the valid solution.
Answer: $\dfrac{1}{\sqrt{3}}$
Hence, the correct answer is option: 'C'.