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Range of Trigonometric Ratios from 0 to 90 Degrees

prove that if...

Question

prove that if x, y ,z belong to R+.
tan(tan^-1x +tan^-1y +tan^-1z)= cot(cot^-1x +cot^-1y +cot^-1z )

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Solution

Dear student,
we know
$a r c \tan (x) = \frac{π}{2} - a r c c o t x u \sin g w e g e t \tan (\tan^{- 1} x + \tan^{- 1} y + \tan^{- 1} z) = \tan (\frac{π}{2} - c o t^{- 1} x + \frac{π}{2} - c o t^{- 1} y + \frac{π}{2} - c o t^{- 1} z) = \tan (\frac{3 π}{2} - c o t^{- 1} x - c o t^{- 1} y - c o t^{- 1} z) = c o t (c o t^{- 1} x + c o t^{- 1} y + c o t^{- 1} z) R e g a r d s$

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Similar questions

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

then prove that:

x + y + z = x y z

If x + y + z = xyz, then tan^-1x + tan^-1y + tan^-1z =

c o t^{- 1} x + c o t^{- 1} y + c o t^{- 1} z = \frac{π}{4}

x y + y z + z x + x + y + z =

c o t^{- 1} x + c o t^{- 1} y + c o t^{- 1} z = \frac{π}{2}, x, y, z > 0 a n d x y < 1, t h e n x + y + z

is also equal to

x + y + z = x y z

x, y, z > 0

, then find the value of

{tan}^{- 1} x + {tan}^{- 1} y + {tan}^{- 1} z

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