If ^ - 1 x + ^ - 1 y + ^ - 1 z = π, then the value of~ k ~in the...

$⇒ \sin^{- 1}x + \sin^{- 1}y = \pi - \sin^{- 1}z$

$⇒ \sin^{- 1}\left[x\sqrt{1 - y^{2}} + y\sqrt{1 - x^{2}}\right] = \pi - \sin^{- 1}z$

$⇒ x\sqrt{1 - y^{2}} + y\sqrt{1 - x^{2}} = z \left(∵ \sin \left(\pi - \theta\right) =sin\theta\right)</p><p>$⇒ $x\sqrt{1 - y^{2}} -$z = - $y\sqrt{1 - x^{2}}</p><p>square both sides we get,~</p><p>$⇒ x^{2} \left(1 - y^{2}\right) + z^{2} - 2xz\sqrt{1 - y^{2}} =y^{2}\left(1 - x^{2}\right)$</p><p>$⇒ x^{2} + z^{2} - y^{2} = 2xz$\sqrt{1 - y^{2}}</p><p>Again square both sides we get,~</p><p>$⇒ x^{4} + z^{4} + y^{2} - 2x^{2}y^{2} + 2x^{2}z^{2} - 2y^{2}z^{2} = 4x^{2}z^{2} - 4x^{2}z^{2}y^{2}$</p><p>$⇒ x^{4} + z^{4} + y^{4} - 4x^{2}z^{2}y^{2} = 2 \left(x^{2}y^{2} + y^{2}z^{2} + z^{2}x^{2}\right)$</p><p>Compare then we get,~$k = 2$.</p>$