If x, y, z are natural numbers such that $c o t^{- 1} x + c o t^{- 1} y = c o t^{- 1} z$ then the number of ordered triplets (x,y,z) that satisfy the equation is ___

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Infinite solutions

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D Infinite solutions
x =1 : y =1 is not a solution of the given equation
Suppose $(x, y) \neq (1, 1)$
Then $\frac{\frac{1}{x} + \frac{1}{y}}{1 - \frac{1}{x y}} = \frac{1}{z} \Rightarrow x (z - y) = - (1 + y z) \Rightarrow= - \frac{(1 + y z)}{z - y}$
If y=n+1:z=n then $x = n^{2} + n + 1$
All such numbers are solutions

Suggest Corrections

Similar questions

c o t^{- 1} x + c o t^{- 1} y = c o t^{- 1} z

c o t^{- 1} x + c o t^{- 1} y = c o t^{- 1} z

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

(x, y, z)

f (x, y, z) = {sin}^{- 1} x + {sin}^{- 1} y + {sin}^{- 1} z = \frac{3 π}{2}

Join BYJU'S Learning Program