Tan-1- radic-x-x-y-z- radic-yz- - tan-1- radic-y-x-y-z- radic-xz- - tan-1- radic-z-x-y-z- radic-xy-0 prove

Dear #160;Student, #160;tan 1xx+y+zyz+tan 1yx+y+zxz+tan 1zx+y+zxy= #160;tan 1xx+y+zyz+yx+y+zxz1 xx+y+zyz #215;yx+y+zxz+tan 1zx+y+zxy= #160; #160;tan ...

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Byju's Answer

Standard XI

Mathematics

Theorems for Differentiability

tan-1[ radi...

Question

tan-1[√x(x+y+z)/√yz] + tan-1[√y(x+y+z)/√xz] + tan-1[√z(x+y+z)/√xy]=0
prove

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Solution

$Dear Student, \tan^{- 1} [\frac{\sqrt{x (x + y + z)}}{\sqrt{yz}}] + \tan^{- 1} [\frac{\sqrt{y (x + y + z)}}{\sqrt{xz}}] + \tan^{- 1} [\frac{\sqrt{z (x + y + z)}}{\sqrt{xy}}] = \tan^{- 1} [\frac{\frac{\sqrt{x (x + y + z)}}{\sqrt{yz}} + \frac{\sqrt{y (x + y + z)}}{\sqrt{xz}}}{1 - \frac{\sqrt{x (x + y + z)}}{\sqrt{yz}} \times \frac{\sqrt{y (x + y + z)}}{\sqrt{xz}}}] + \tan^{- 1} [\frac{\sqrt{z (x + y + z)}}{\sqrt{xy}}] = \tan^{- 1} [\frac{\sqrt{x^{2} (x + y + z)} + \sqrt{y^{2} (x + y + z)}}{\sqrt{xyz} [\frac{z - x - y - z}{z}]}] + \tan^{- 1} [\frac{\sqrt{z (x + y + z)}}{\sqrt{xy}}] = \tan^{- 1} [- \frac{z [\sqrt{x^{2} (x + y + z)} + \sqrt{y^{2} (x + y + z)}]}{\sqrt{xyz} [x + y]}] + \tan^{- 1} [\frac{\sqrt{z (x + y + z)}}{\sqrt{xy}}] = \tan^{- 1} [\frac{\sqrt{z (x + y + z)}}{\sqrt{xy}}] - \tan^{- 1} [\frac{z \sqrt{(x + y + z)} [x + y]}{\sqrt{xyz} [x + y]}] = \tan^{- 1} [\frac{\sqrt{z (x + y + z)}}{\sqrt{xy}}] - \tan^{- 1} [\frac{\sqrt{z (x + y + z)}}{\sqrt{xy}}] = 0 Proved . Regards$

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Similar questions

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

= π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

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

Q. If

x > 0, y > 0, z > 0, x y + y z + z x < 1

and if

{tan}^{- 1} + {tan}^{- 1} y + {tan}^{- 1} z = π

then

x + y + z =

Q. If

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

, then prove that,

x y + y z + z x = 1

If $t a n^{- 1} x + t a n^{- 1} y + t a n^{- 1} z = \frac{π}{2}$ then prove that $x y + y z + z x = 1$ .

$t a n^{- 1} x + t a n^{- 1} y + t a n^{- 1} z = \frac{π}{2} \Rightarrow 1 - x y - y z - z x =$

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tan-1[√x(x+y+z)/√yz] + tan-1[√y(x+y+z)/√xz] + tan-1[√z(x+y+z)/√xy]=0 prove

Dear Student, tan-1xx+y+zyz+tan-1yx+y+zxz+tan-1zx+y+zxy= tan-1xx+y+zyz+yx+y+zxz1-xx+y+zyz×yx+y+zxz+tan-1zx+y+zxy= tan-1x2x+y+z+y2x+y+zxyzz-x-y-zz+tan-1zx+y+zxy= tan-1-zx2x+y+z+y2x+y+zxyzx+y+tan-1zx+y+zxy=tan-1zx+y+zxy-tan-1zx+y+zx+yxyzx+y=tan-1zx+y+zxy-tan-1zx+y+zxy = 0 Proved.Regards

tan-1[√x(x+y+z)/√yz] + tan-1[√y(x+y+z)/√xz] + tan-1[√z(x+y+z)/√xy]=0
prove