The value of $2\tan^{-1}\sqrt{x} - \cos^{-1}\left(\dfrac{1-x}{1+x}\right)$ is (a) $0$ (b) $1$ (c) $\dfrac{1}{3}$ (d) $\dfrac{1}{2}$

The value of 2tan⁻¹√(x) - cos⁻¹((1-x)/(1+x)) is (a – Class 12 Mathematics

Question 42 of 43

2\tan^{-1}\sqrt{x} - \cos^{-1}\left(\dfrac{1-x}{1+x}\right)

(a)

0

(b)

1

(c)

\dfrac{1}{3}

(d)

\dfrac{1}{2}

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2026MCQ· 1mImportance★★★★★

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Substitute $\sqrt{x} = \tan\theta$ ; then $\dfrac{1-x}{1+x} = \cos 2\theta$ and both terms equal $2\theta$ , so the difference is $0$ .

This inverse-trig simplification mirrors the NCERT/CBSE Class 12 technique of the $\tan\theta$ substitution.

Let $\sqrt{x} = \tan\theta$ with $\theta \in [0, \tfrac{\pi}{2})$ , so $x = \tan^2\theta$ and $\tan^{-1}\sqrt{x} = \theta$ . Then

$\frac{1-x}{1+x} = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta.$

Since $2\theta \in [0,\pi)$ , the principal value gives …