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If-xcos-ycos-zcos-0-xsin-ysin-zsin-0-and-xsec-ysec-zsec-0-then-prove-that-x-2-y-2-z-2-2-4x-2-y-2-




Question Number 157031 by PRITHWISH SEN 2 last updated on 18/Oct/21
If  xcos θ+ycos ∅+zcos ψ=0,        xsin θ+ysin ∅+zsin ψ=0  and xsec θ+ysec ∅+zsec ψ=0  then prove that            (x^2 +y^2 −z^2 )^2 = 4x^2 y^2
$$\mathrm{If}\:\:\mathrm{xcos}\:\theta+\mathrm{ycos}\:\emptyset+\mathrm{zcos}\:\psi=\mathrm{0}, \\ $$$$\:\:\:\:\:\:\mathrm{xsin}\:\theta+\mathrm{ysin}\:\emptyset+\mathrm{zsin}\:\psi=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{xsec}\:\theta+\mathrm{ysec}\:\emptyset+\mathrm{zsec}\:\psi=\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{z}^{\mathrm{2}} \right)^{\mathrm{2}} =\:\mathrm{4x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \\ $$

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