If ^ -1 x + ^ -1 x = π / 2 , then 2x^ 2 + 1 =

We have,

$\sin^{-1}x + \tan^{-1}x = \dfrac{\pi}{2} ⇒ 0 < x < 1$

$∴ \tan^{-1}x = \dfrac{\pi}{2} - \sin^{-1}x = \cos^{-1}x$

$⇒ \tan^{-1}x = \tan^{-1} \left(\dfrac{1 - x^{2}}{x}\right)</p><p>$⇒ x^{2} = $\sqrt{1 - x^{2}}</p><p>$⇒ x^{4} + x^{2} - 1 = 0 ⇒ x^{2} = $\dfrac{- 1 \pm \sqrt{5}}{2}</p><p>as$ x^{2} > 0$</p><p>$∴ x^{2} = $\dfrac{\sqrt{5} - 1}{2}</p><p>$⇒ 2x^{2} + 1 = $\sqrt{5}$