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Question Number 196883 by Amidip last updated on 02/Sep/23

	Answered by som(math1967) last updated on 02/Sep/23
	$sin^(−1) x+sin^(−1) y=π−sin^(−1) z ⇒sin^(−1) (x(√(1−y^2 ))+y(√(1−x^2 )))=sin^(−1) z ⇒x(√(1−y^2 ))−z=−y(√(1−x^2 )) ⇒{x(√(1−y^2 ))−z}^2 =y^2 −x^2 y^2 ⇒x^2 −x^2 y^2 +z^2 −2xz(√(1−y^2 ))=y^2 −x^2 y^2 ⇒(x^2 −y^2 +z^2 )^2 =4x^2 z^2 −4x^2 y^2 z^2 ⇒(2xz)^2 −(x^2 −y^2 +z^2 )^2 =4x^2 y^2 z^2 ⇒(2xz+x^2 +z^2 −y^2 )(y^2 −x^2 +2xz−z^2 ) =4x^2 y^2 z^2 {(x+z)^2 −y^2 }{y^2 −(x−z)^2 }=4x^2 y^2 z^2 (x+y+z)(x+z−y)(y−x+z)(x+y−z) =4x^2 y^2 z^2$
	$\sin^{- 1} x + \sin^{- 1} y = π - \sin^{- 1} z$ $\Rightarrow {s i n}^{- 1} (x \sqrt{1 - y^{2}} + y \sqrt{1 - x^{2}}) = \sin^{- 1} z$ $\Rightarrow x \sqrt{1 - y^{2}} - z = - y \sqrt{1 - x^{2}}$ $\Rightarrow {x \sqrt{1 - y^{2}} - z}^{2} = y^{2} - x^{2} y^{2}$ $\Rightarrow x^{2} - x^{2} y^{2} + z^{2} - 2 x z \sqrt{1 - y^{2}} = y^{2} - x^{2} y^{2}$ $\Rightarrow {(x^{2} - y^{2} + z^{2})}^{2} = 4 x^{2} z^{2} - 4 x^{2} y^{2} z^{2}$ $\Rightarrow {(2 x z)}^{2} - {(x^{2} - y^{2} + z^{2})}^{2} = 4 x^{2} y^{2} z^{2}$ $\Rightarrow (2 x z + x^{2} + z^{2} - y^{2}) (y^{2} - x^{2} + 2 x z - z^{2})$ $= 4 x^{2} y^{2} z^{2}$ ${{(x + z)}^{2} - y^{2}} {y^{2} - {(x - z)}^{2}} = 4 x^{2} y^{2} z^{2}$ $(x + y + z) (x + z - y) (y - x + z) (x + y - z)$ $= 4 x^{2} y^{2} z^{2}$
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