Question

If sin^–1x + sin^–1y + sin^–1z = π, show that `x^2 - y^2 - z^2 + 2yzsqrt(1 - x^2) = 0`

Answer 1

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