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Question: If tan<sup>–1</sup> x + tan<sup>–1</sup> y + tan<sup>–1</sup> z = π, then\(\frac{1}{xy}\)+ \(\frac{1...

If tan–1 x + tan–1 y + tan–1 z = π, then1xy\frac{1}{xy}+ 1yz\frac{1}{yz}+ 1zx\frac{1}{zx}=

A

0

B

1

C

1xyz\frac{1}{xyz}

D

xyz

Answer

1

Explanation

Solution

tan–1 x + tan–1 y + tan–1 z = π

⇒ tan–1 [x+y+z−xyz1−(xy+yz+zx)]\left[ \frac { x + y + z - x y z } { 1 - ( x y + y z + z x ) } \right]= π

⇒ x+y+z−xyz1−(xy+yz+zx)\frac { x + y + z - x y z } { 1 - ( x y + y z + z x ) }= tan π = 0

⇒ x + y + z – xyz = 0⇒ ++ 1zx\frac { 1 } { \mathrm { zx } } = 1