Question

if sin inverse x +sin inverse y +sin inverse z = pi the prove that x^2-y^2-z^2+2yzsqrt(1-x^2) =0​

Answer 1

Step-by-step explanation:

We have,

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

$\implies \sin^{ - 1} (x) = \pi - \sin^{ - 1} (y) - \sin^{ - 1} (z)$

$\implies \sin^{ - 1} (x) = \frac{\pi}{2} - \sin^{ - 1} (y) + \frac{\pi}{2} - \sin^{ - 1} (z)$

$\implies \sin^{ - 1} (x) = \cos^{ - 1} (y) + \cos^{ - 1} (z)$

$\implies \cos \{\sin^{ - 1} (x) \} = \cos \{ \cos^{ - 1} (y) + \cos^{ - 1} (z) \}$

$\implies \cos \{\cos^{ - 1} ( \sqrt{1 - {x}^{2} } ) \} = \cos \{ \cos^{ - 1} (y) \} \cos \{ \cos^{ - 1} (z) \} - \sin \{ \cos^{ -1 } (y) \}\sin \{ \cos^{ - 1} (z) \}$

$\implies \sqrt{1 - {x}^{2} } =yz - \sin \{ \sin^{ -1 } ( \sqrt{1 - {y}^{2} } ) \}\sin \{ \sin^{ - 1} ( \sqrt{1 - z^{2} } ) \}$

$\implies \sqrt{1 - {x}^{2} } =yz - \sqrt{1 - {y}^{2} } . \sqrt{1 - z^{2} }$

$\implies \sqrt{1 - {y}^{2} } . \sqrt{1 - z^{2} } = yz - \sqrt{1 - {x}^{2} }$

$\implies( \sqrt{1 - {y}^{2} } . \sqrt{1 - z^{2} } )^{2} = (yz - \sqrt{1 - {x}^{2} } )^{2}$

$\implies(1 - {y}^{2}) .(1 - z^{2} )= y^{2} z^{2} + 1 - {x}^{2} - 2yz \sqrt{1 - {x}^{2} }$

$\implies1 - {z}^{2} - {y}^{2} + {y}^{2} {z}^{2} = y^{2} z^{2} + 1 - {x}^{2} - 2yz \sqrt{1 - {x}^{2} }$

$\implies - {z}^{2} - {y}^{2} = - {x}^{2} - 2yz \sqrt{1 - {x}^{2} }$

$\implies {x}^{2} - {z}^{2} - {y}^{2} = - 2yz \sqrt{1 - {x}^{2} }$

$\implies {x}^{2} - {z}^{2} - {y}^{2} + 2yz \sqrt{1 - {x}^{2} } = 0$

Hence, proved