Question

If \[3\sin^{- 1} \left( \frac{2x}{1 + x^2} \right) - 4 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) + 2 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) = \frac{\pi}{3}\] is equal to

 

विकल्प

• `1/sqrt3`

• `-1/sqrt3`

• `sqrt3`

• `-sqrt3/4`

Answer 1

(a) `1/sqrt3`

Let `x=tany`
Then,
\[3 \sin^{- 1} \left( \frac{2\tan{y}}{1 + \tan^2 y} \right) - 4\left( \frac{1 - \tan^2 y}{1 + \tan^2 y} \right) + 2 \tan^{- 1} \left( \frac{2\tan{y}}{1 - \tan^2 y} \right) = \frac{\pi}{3}\]
\[ \Rightarrow 3 \sin^{- 1} \left( \sin 2y \right) - 4 \cos^{- 1} \left( \cos 2y \right) + 2 \tan^{- 1} \left( \tan2y \right) = \frac{\pi}{3} \]
\[ \left[ \because \sin2y = \left( \frac{2\tan{y}}{1 + \tan^2 y} \right), \cos2y = \left( \frac{1 - \tan^2 y}{1 + \tan^2 y} \right) \text{ and }\tan2y = \left( \frac{2\tan{y}}{1 - \tan^2 y} \right) \right]\]
\[ \Rightarrow 3 \times 2y - 4 \times 2y + 2 \times 2y = \frac{\pi}{3}\]
\[ \Rightarrow 6y - 8y + 4y = \frac{\pi}{3}\]
\[ \Rightarrow 2y = \frac{\pi}{3}\]
\[ \Rightarrow y = \frac{\pi}{6}\]
\[ \Rightarrow \tan^{- 1} x = \frac{\pi}{6} \left[ \because \tan^{- 1} x = y \right]\]
\[ \Rightarrow x = \tan\frac{\pi}{6}\]
\[ \Rightarrow x = \frac{1}{\sqrt{3}}\]