The value of tan -1(√(x(x+y+z)/y z))+ tan -1(√(y(x+y+z)/z x))+ tan -1(√(z(x+y+z)/x y)) is (where x , y , z are distinct positive real numbers)

Question

The value of tan -1(√(x(x+y+z)/y z))+ tan -1(√(y(x+y+z)/z x))+ tan -1(√(z(x+y+z)/x y)) is (where x , y , z are distinct positive real numbers) (A)  π (B) (π/4) (C) (π/2) (D) (3 π/4)

Tardigrade Admin · Accepted Answer

Let a=√(x+y+z/x y z) tan -1(a x)+ tan -1(a y)+ tan -1(a z)= tan -1((a x+a y+a z-a3 x y z/1-())) = tan -1((a(x+y+z-a2 x y z)/d r))= tan -1((0/d r))=π .

Q. The value of $tan^{- 1} (\frac{x ( x + y + z )}{yz}) + tan^{- 1} (\frac{y ( x + y + z )}{z x}) + tan^{- 1} (\frac{z ( x + y + z )}{x y})$ is (where $x, y, z$ are distinct positive real numbers)

$π$

$\frac{π}{4}$

$\frac{π}{2}$

$\frac{3 π}{4}$

Let $a = \frac{x + y + z}{x yz}$
$tan^{- 1} (a x) + tan^{- 1} (a y) + tan^{- 1} (a z) = tan^{- 1} (\frac{a x + a y + a z - a ^{3} x yz}{1 - ( )})$
$= tan^{- 1} (\frac{a ( x + y + z - a ^{2} x yz )}{d r}) = tan^{- 1} (\frac{0}{d r}) = π$ .

Q. The value of tan−1(yzx(x+y+z)​​)+tan−1(zxy(x+y+z)​​)+tan−1(xyz(x+y+z)​​) is (where x,y,z are distinct positive real numbers)

π

4π​

2π​

43π​

Let a=xyzx+y+z​​ tan−1(ax)+tan−1(ay)+tan−1(az)=tan−1(1−()ax+ay+az−a3xyz​) =tan−1(dra(x+y+z−a2xyz)​)=tan−1(dr0​)=π .

Q. The value of $tan^{- 1} (\frac{x ( x + y + z )}{yz}) + tan^{- 1} (\frac{y ( x + y + z )}{z x}) + tan^{- 1} (\frac{z ( x + y + z )}{x y})$ is (where $x, y, z$ are distinct positive real numbers)

$π$

$\frac{π}{4}$

$\frac{π}{2}$

$\frac{3 π}{4}$

Let $a = \frac{x + y + z}{x yz}$
$tan^{- 1} (a x) + tan^{- 1} (a y) + tan^{- 1} (a z) = tan^{- 1} (\frac{a x + a y + a z - a ^{3} x yz}{1 - ( )})$
$= tan^{- 1} (\frac{a ( x + y + z - a ^{2} x yz )}{d r}) = tan^{- 1} (\frac{0}{d r}) = π$ .