Question

if cot inverse x+cot inverse y+cot inverse z=π/2 then what is the value of x+y+z​

Answer 1

${cot}^{ - 1} x + {cot}^{ - 1} y + {cot}^{ - 1} z = \frac{\pi}{2} \\ \\ taking \: cotangents \: of \: both \: the \: sides \\ \\ cot( {cot}^{ - 1} x + {cot}^{ - 1} y + {cot}^{ - 1} z) \\ = cot \frac{\pi}{2} \\ cot. {cot}^{ - 1} x + cot. {cot}^{ - 1} y + cot. {cot}^{ - 1} z \\ = 0 \\ (1)x + (1)y + (1)z = 0 \\ x + y + z = 0$

Hence, x + y + z = 0