Question

If `tan^-1x + tan^-1y + tan^-1z = pi/2`, then

पर्याय

• x + y + z – xyz = 0

• xy + yz + zx – 1 = 0

• x + y + z + xyz = 0

• xy + yz + zx + 1 = 0

Answer 1

xy + yz + zx – 1 = 0

Explanation:

`tan^-1x + tan^-1y + tan^-1z = pi/2`

⇒ `tan^-1x + tan^-1y = pi/2 - tan^-1z`

⇒ `tan^-1 ((x + y)/(1 - xy)) = cot^-1z`

⇒ `((x + y)/(1 - xy)) = tan(tan^-1  1/z)`

⇒ `z(x + y) = 1 - xy`

⇒ `xy + yz + zx - 1` = 0