To prove: ${\mathrm{cot}}^{-1}7+{\mathrm{cot}}^{-1}8+{\mathrm{cot}}^{-1}18={\mathrm{cot}}^{-1}3$

Jace Wright
2022-05-19
Answered

To prove: ${\mathrm{cot}}^{-1}7+{\mathrm{cot}}^{-1}8+{\mathrm{cot}}^{-1}18={\mathrm{cot}}^{-1}3$

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Blaine Andrews

Answered 2022-05-20
Author has **20** answers

The statement is equivalent to

$\mathrm{arccot}7+\mathrm{arccot}8=\mathrm{arccot}3-\mathrm{arccot}18$

The right-hand side is positive because the arccotangent is decreasing, so it is in the interval $(0,\pi /2)$ and we have to ensure that the same holds for the left-hand side, but this is easy, because 7>1 and 8>1, hence

$\mathrm{arccot}7+\mathrm{arccot}8<2\mathrm{arccot}1=\frac{\pi}{2}$

The two sides are equal if and only if their cotangents are.

Since

$\mathrm{cot}(\alpha +\beta )=\frac{\mathrm{cot}\alpha \mathrm{cot}\beta -1}{\mathrm{cot}\alpha +\mathrm{cot}\beta}\phantom{\rule{2em}{0ex}}\mathrm{cot}(\alpha -\beta )=\frac{\mathrm{cot}\alpha \mathrm{cot}\beta +1}{\mathrm{cot}\beta -\mathrm{cot}\alpha}$

this amounts to proving that

$\frac{7\cdot 8-1}{7+8}=\frac{3\cdot 18+1}{18-3}$

$\mathrm{arccot}7+\mathrm{arccot}8=\mathrm{arccot}3-\mathrm{arccot}18$

The right-hand side is positive because the arccotangent is decreasing, so it is in the interval $(0,\pi /2)$ and we have to ensure that the same holds for the left-hand side, but this is easy, because 7>1 and 8>1, hence

$\mathrm{arccot}7+\mathrm{arccot}8<2\mathrm{arccot}1=\frac{\pi}{2}$

The two sides are equal if and only if their cotangents are.

Since

$\mathrm{cot}(\alpha +\beta )=\frac{\mathrm{cot}\alpha \mathrm{cot}\beta -1}{\mathrm{cot}\alpha +\mathrm{cot}\beta}\phantom{\rule{2em}{0ex}}\mathrm{cot}(\alpha -\beta )=\frac{\mathrm{cot}\alpha \mathrm{cot}\beta +1}{\mathrm{cot}\beta -\mathrm{cot}\alpha}$

this amounts to proving that

$\frac{7\cdot 8-1}{7+8}=\frac{3\cdot 18+1}{18-3}$

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The equation is in one dimension and P and O are points on the x axis.

The Electric field is oriented in the $+x$ direction ans has the formula $c\frac{1}{r}$ where c is a coefficient and r is the distance form the origin of the coordinate system.

The vector ${\overrightarrow{r}}_{P}-{\overrightarrow{r}}_{O}$ has the same direction as that of the electric field.

Now, Why do we put P below the integral and O above it? Can someone explain the equation for me?

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