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प्रश्न
If sin–1x + sin–1y + sin–1z = π, show that `x^2 - y^2 - z^2 + 2yzsqrt(1 - x^2) = 0`
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उत्तर
Given, sin–1x + sin–1y + sin–1z = π
`\implies` sin–1x + sin–1y = π – sin–1z
`\implies sin^-1[xsqrt(1 - y^2) + ysqrt(1 - x^2)] = (pi - sin^-1z)`
`\implies xsqrt(1 - y^2) + ysqrt(1 - x^2) = sin(pi - sin^-1z)`
`\implies xsqrt(1 - y^2) + ysqrt(1 - x^2) = z`
`\implies xsqrt(1 - y^2) = z - ysqrt(1 - x^2)`
Now squaring on both sides, we get,
`(xsqrt(1 - y^2))^2 = (z - ysqrt(1 - x^2))^2`
`\implies x^2(1 - y^2) = (z^2 + y^2(1 - x^2) - 2zy sqrt(1 - x^2))`
`\implies x^2 - x^2y^2 = z^2 + y^2 - x^2y^2 - 2yz sqrt(1 - x^2)`
`\implies x^2 - y^2 - z^2 + 2yz sqrt(1 - x^2)` = 0
Hence proved
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