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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

$$\:\:\mathrm{sin}^{−\mathrm{1}} {x}+\mathrm{sin}^{−\mathrm{1}} {y}=\pi−\mathrm{sin}^{−\mathrm{1}} {z} \\ $$$$\Rightarrow{sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }+{y}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)=\mathrm{sin}^{−\mathrm{1}} {z} \\ $$$$\Rightarrow{x}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }−{z}=−{y}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\left\{{x}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }−{z}\right\}^{\mathrm{2}} ={y}^{\mathrm{2}} −{x}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{xz}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }={y}^{\mathrm{2}} −{x}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{4}{x}^{\mathrm{2}} {z}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} {y}^{\mathrm{2}} {z}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{2}{xz}\right)^{\mathrm{2}} −\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{4}{x}^{\mathrm{2}} {y}^{\mathrm{2}} {z}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{2}{xz}+{x}^{\mathrm{2}} +{z}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)\left({y}^{\mathrm{2}} −{x}^{\mathrm{2}} +\mathrm{2}{xz}−{z}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:=\mathrm{4}{x}^{\mathrm{2}} {y}^{\mathrm{2}} {z}^{\mathrm{2}} \\ $$$$\left\{\left({x}+{z}\right)^{\mathrm{2}} −{y}^{\mathrm{2}} \right\}\left\{{y}^{\mathrm{2}} −\left({x}−{z}\right)^{\mathrm{2}} \right\}=\mathrm{4}{x}^{\mathrm{2}} {y}^{\mathrm{2}} {z}^{\mathrm{2}} \\ $$$$\left({x}+{y}+{z}\right)\left({x}+{z}−{y}\right)\left({y}−{x}+{z}\right)\left({x}+{y}−{z}\right) \\ $$$$\:=\mathrm{4}{x}^{\mathrm{2}} {y}^{\mathrm{2}} {z}^{\mathrm{2}} \\ $$

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