-1x y-1-x-y- -1y z-1-y-z- -1x z-1-z-x is equal to

The correct option is D 0cot^ 1xy+1/x y+cot^ 1yz+1/y z+cot^ 1xz+1/z x =tan^ 1x y/1+xy+tan^ 1y z/1+yz+cot^ 1xz+1/z x =(tan^ 1x tan^ 1y)+(tan^ 1y tan^ 1z)+(tan^ ...

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Byju's Answer

Standard XII

Mathematics

Property 1

-1x y+1/x-y+ ...

Question

$c o t^{- 1} \frac{x y + 1}{x - y} + c o t^{- 1} \frac{y z + 1}{y - z} + c o t^{- 1} \frac{x z + 1}{z - x}$ is equal to

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Solution

The correct option is C
0

$c o t^{- 1} \frac{x y + 1}{x - y} + c o t^{- 1} \frac{y z + 1}{y - z} + c o t^{- 1} \frac{x z + 1}{z - x}$
$= t a n^{- 1} \frac{x - y}{1 + x y} + t a n^{- 1} \frac{y - z}{1 + y z} + c o t^{- 1} \frac{x z + 1}{z - x}$
$= (t a n^{- 1} x - t a n^{- 1} y) + (t a n^{- 1} y - t a n^{- 1} z) + (t a n^{- 1} z - t a n^{- 1} x) = 0$

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Similar questions

{cot}^{- 1} \frac{x y + 1}{x - y} + {cot}^{1} \frac{y z + 1}{y - z} + {cot}^{- 1} \frac{x z + 1}{z - x}

Q. Solve:

{cot}^{- 1} (\frac{x y + 1}{x - y}) + {cot}^{- 1} (\frac{y z + 1}{y - z}) + {cot}^{- 1} (\frac{z x + 1}{z - x})

Q. If

x > y > z > 0

, then find the value of

{cot}^{- 1} \frac{x y + 1}{x - y} + {cot}^{- 1} \frac{y z + 1}{y - z} + {cot}^{- 1} \frac{z x + 1}{z - x}

Q. If

x^{2} + y^{2} + z^{2} = r^{2}

, then

{tan}^{- 1} (\frac{x y}{z r}) + {tan}^{- 1} (\frac{y z}{x r}) + {tan}^{- 1} (\frac{x z}{y r}) =

Q. If [a] denotes the greatest integer less than or equal to a and

- 1 \leq x < 0, 0 \leq y < 1, 1 \leq z < 2

, then

∣ ∣ ∣ ∣ ∣ \begin{matrix} [x] + 1 & [y] & [z] [x] & [y] + 1 & [z] [x] & [y] & [z] + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

is equal to

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cot−1xy+1x−y+cot−1yz+1y−z+cot−1xz+1z−x is equal to

The correct option is C 0cot−1xy+1x−y+cot−1yz+1y−z+cot−1xz+1z−x =tan−1x−y1+xy+tan−1y−z1+yz+cot−1xz+1z−x =(tan−1x−tan−1y)+(tan−1y−tan−1z)+(tan−1z−tan−1x)=0

$c o t^{- 1} \frac{x y + 1}{x - y} + c o t^{- 1} \frac{y z + 1}{y - z} + c o t^{- 1} \frac{x z + 1}{z - x}$ is equal to