If tan-1x-tan-1y-tan-1z- then 1-xy-1-yz-1-zx-

Tan^ 1(x)+tan^ 1(y)+tan^ 1(z)=π ⇒ tan^ 1x+tan^ 1y=π tan^ 1z ⇒x+y/1 xy= z⇒ x+y= z+xyz ⇒ x+y+z=xyz Dividing by xyz, we get 1/yz+1/xz+1/xy=1. Note: Students should ...

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Standard XI

Mathematics

Properties Derived from Trigonometric Identities

If tan-1x+tan...

Question

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

0

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1

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\frac{1}{x y z}

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x y z

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Solution

The correct option is B $1$
$t a n^{- 1} (x) + t a n^{- 1} (y) + t a n^{- 1} (z) = π$
$\Rightarrow t a n^{- 1} x + t a n^{- 1} y = π - t a n^{- 1} z$
$\Rightarrow \frac{x + y}{1 - x y} = - z \Rightarrow x + y = - z + x y z$
$\Rightarrow x + y + z = x y z$
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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Similar questions

$t a n^{- 1} x + t a n^{- 1} y + t a n^{- 1} z = \frac{π}{2} \Rightarrow 1 - x y - y z - z x =$

If $t a n^{- 1} x + t a n^{- 1} y + t a n^{- 1} z = \frac{π}{2}$ then prove that $x y + y z + z x = 1$ .

Q. If

x y + y z + z x = 1

then find the value of

{tan}^{- 1} x + {tan}^{- 1} y + {tan}^{- 1} z

Q. 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} =

Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

(a) If

{c o s}^{- 1} p + {c o s}^{- 1} q + {c o s}^{- 1} r = π

, then prove that

p^{2} + q^{2} + r^{2} + 2 p q r = 1

(b) If

{s i n}^{- 1} x + {s i n}^{- 1} y + {s i n}^{- 1} z = π

, then prove that

x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = 2 (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})

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

π / 2

show that

x + y + z = x y z

x y + y z + z x = 1

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

The correct option is B 1tan−1(x)+tan−1(y)+tan−1(z)=π ⇒tan−1x+tan−1y=π−tan−1z ⇒x+y1−xy=−z⇒x+y=−z+xyz ⇒x+y+z=xyz Dividing by xyz, we get 1yz+1xz+1xy=1. Note: Students should remember this question as a formula.

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