Solution of the equation cos−1⁡x2−1x2+1+sin−1⁡2xx2+1+tan−1⁡2xx2−1=2π3 is / are

L . H . S = cos − 1 ⁡ − 1 − x 2 1 + x 2 + sin − 1 ⁡ 2 x 1 + x 2 + tan − 1 ⁡ − 2 x 1 − x 2 = π − cos − 1 ⁡ 1 − x 2 1 + x 2 + sin − 1 ⁡ 2 x 1 + x 2 − tan − 1 ⁡ 2 x 1 − x 2 Case I : x< -1 = π + 2 tan − 1 ⁡ x + − π − 2 tan − 1 ⁡ x + π + 2 tan − 1 ⁡ x = 2 π 3 ⇒ 2 tan − 1 ⁡ x = 2 π 3 − π = − π 3 ⇒ x = tan ⁡ − π 6 = − 1 3 Since − 1 3 > − 1 so x = − 1 3 is not admissible. Case II :-1 < x <0 = π + 2 tan − 1 ⁡ x + 2 tan − 1 ⁡ x − 2 tan − 1 ⁡ x = 2 π 3 ⇒ tan − 1 ⁡ x = − π 6 ⇒ x = − tan ⁡ π 6 = − 1 3 (Valid) Case III : 0 < x < 1 = π − 2 tan − 1 ⁡ x + 2 tan − 1 ⁡ x − 2 tan − 1 ⁡ x = 2 π 3 ⇒ tan − 1 ⁡ x = π 6 ⇒ x = 1 3 valid Case IV : x>1 L.H.S. = π − 2 tan − 1 ⁡ x + π − 2 tan − 1 ⁡ x + π − 2 tan − 1 ⁡ x = 2 π 3 ⇒ − 6 tan − 1 ⁡ x = 2 π 3 − 3 π = − 7 π 3 ⇒ tan − 1 ⁡ x = 7 π 18 ⇒ x = tan ⁡ π 2 − π 9 = cot ⁡ π 9 = cot ⁡ 20 ∘ > 1

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Solution of the equation $\cos^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + \sin^{- 1} (\frac{2 x}{x^{2} + 1}) + \tan^{- 1} (\frac{2 x}{x^{2} - 1}) = \frac{2 π}{3}$ is / are

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tan20^o

$- \frac{1}{\sqrt{3}}$

$\frac{1}{\sqrt{3}}$

cot20^o

answer is A, B, D.

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Detailed Solution

$\begin{aligned} L . H . S = \cos^{- 1} (- \frac{1 - x^{2}}{1 + x^{2}}) + \sin^{- 1} \frac{2 x}{1 + x^{2}} + \tan^{- 1} (- \frac{2 x}{1 - x^{2}}) \\ = π - \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + \sin^{- 1} \frac{2 x}{1 + x^{2}} - \tan^{- 1} \frac{2 x}{1 - x^{2}} \end{aligned}$
Case I : x< -1
$= π + 2 \tan^{- 1} x + (- π - 2 \tan^{- 1} x) + (π + 2 \tan^{- 1} x)$
$\begin{array}{l} = \frac{2 π}{3} \\ \Rightarrow 2 \tan^{- 1} x = \frac{2 π}{3} - π = - \frac{π}{3} \Rightarrow x = \tan (- \frac{π}{6}) = - \frac{1}{\sqrt{3}} \\ Since - \frac{1}{\sqrt{3}} > - 1 so x = - \frac{1}{\sqrt{3}} is not admissible. \end{array}$
Case II :-1 < x <0
$= π + 2 \tan^{- 1} x + 2 \tan^{- 1} x - 2 \tan^{- 1} x = \frac{2 π}{3}$
$\Rightarrow \tan^{- 1} x = - \frac{π}{6} \Rightarrow x = - \tan \frac{π}{6} = - \frac{1}{\sqrt{3}} (Valid)$
Case III : 0 < x < 1
$= π - 2 \tan^{- 1} x + 2 \tan^{- 1} x - 2 \tan^{- 1} x = \frac{2 π}{3}$
$\Rightarrow \tan^{- 1} x = \frac{π}{6} \Rightarrow x = \frac{1}{\sqrt{3}} (valid)$
Case IV : x>1
$\begin{array}{l} L.H.S. = π - 2 \tan^{- 1} x + π - 2 \tan^{- 1} x + π - 2 \tan^{- 1} x = \frac{2 π}{3} \\ \Rightarrow - 6 \tan^{- 1} x = \frac{2 π}{3} - 3 π = - \frac{7 π}{3} \\ \Rightarrow \tan^{- 1} x = \frac{7 π}{18} \Rightarrow x = \tan (\frac{π}{2} - \frac{π}{9}) = \cot \frac{π}{9} = \cot 20^{\circ} > 1 \end{array}$