If x2-y2-z2-r2- then tan-1xy-zr-tan-1yz-xr-tan-1xz-yr-

The correct option is A π/2 tan ^ 1A+tan ^ 1B+tan ^ 1C=tan ^ 1(A+B+C ABC1 AB BC AC) tan ^ 1(xyzr)+tan ^ 1(yzxr)+tan ^ 1(xzyr) =tan ^ 1( xyz ...

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Definite Integral as Limit of Sum

If x2+y2+z2...

Question

If $x^{2} + y^{2} + z^{2} = r^{2}$ , then ${tan}^{- 1} (\frac{x y}{z r}) + {tan}^{- 1} (\frac{y z}{x r}) + {tan}^{- 1} (\frac{x z}{y r}) =$

π

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\frac{π}{2}

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0

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\frac{π}{4}

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Solution

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r^{2} = x^{2} + y^{2} + z^{2}

{tan}^{- 1} \frac{y z}{x r} + {tan}^{- 1} \frac{x z}{y r} = \frac{π}{2} - {tan}^{- 1} ϕ

x, y, z

r

x^{2} + y^{2} + z^{2} = r^{2} .

{tan}^{- 1} (\frac{x y}{z r}) + {tan}^{- 1} (\frac{y z}{x r}) + tan (\frac{z x}{y r})

x^{2} + y^{2} + z^{2} = r^{2}

tan α = \frac{x y}{z r}, tan β = \frac{y z}{x r}, tan γ = \frac{Z X}{y r}

α + β + γ =

\sum t a n^{- 1} (\frac{y z}{x \sqrt{x^{2} + y^{2} + z^{2}}}) = ?

r^{2} = x^{2} + y^{2} + z^{2}

{tan}^{- 1} (\frac{y z}{r x}) + {tan}^{- 1} (\frac{z x}{r y}) + {tan}^{- 1} (\frac{x y}{r z}) = \frac{π}{2}

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