If t a n − 1 ( x ) + t a n − 1 ( y ) + t a n − 1 ( z ) = π , then 1 x y + 1 y z + 1 z x = Options: A . &nbsp 0 B . &nbsp 1 C . &nbsp 1 x y z D . &nbsp x y z

Answer: Option B : B t a n − 1 ( x ) + t a n − 1 ( y ) + t a n − 1 ( z ) = π ⇒ t a n − 1 x + t a n − 1 y = π − t a n − 1 z ⇒ x + y 1 − x y = − z ⇒ x + y = − z + x y z ⇒ x + y + z = x y z Dividing by xyz, we get 1 y z + 1 x z + 1 x y = 1. Note: Students should remember this question as a formula.

If tan−1(x)+tan−1(y)+tan−1(z)=π, then 1xy+1yz+1zx=

Discussion Forum : Inverse Trigonometric Functions

Question -

If $t a n^{- 1} (x) + t a n^{- 1} (y) + t a n^{- 1} (z) = π$ , then $\frac{1}{x y} + \frac{1}{y z} + \frac{1}{z x} =$

Options:

A .

0

B .

1

C .

\frac{1}{x y z}

D .

x y z

Answer: Option B
:
B

t a n^{- 1} (x) + t a n^{- 1} (y) + t a n^{- 1} (z) = π

\Rightarrow t a n^{- 1} x + t a n^{- 1} y = π - t a n^{- 1} z

\Rightarrow \frac{x + y}{1 - x y} = - z \Rightarrow x + y = - z + x y z

\Rightarrow x + y + z = x y z

Dividing by xyz, we get

\frac{1}{y z} + \frac{1}{x z} + \frac{1}{x y} = 1.

Note: Students should remember this question as a formula.

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Discussion Forum : Inverse Trigonometric Functions

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