If $x = {tan}^{- 1} t$ $x=tan^(-1)t$ , then is $\frac{1}{x} = \frac{1}{tan t}$ $1/x=1/tant$ ?

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If $x = {tan}^{- 1} t$ $x=tan^(-1)t$ , then is $\frac{1}{x} = \frac{1}{tan t}$ $1/x=1/tant$ ?

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Answer 1 · 2017-02-08T09:28:22

$\frac{1}{x} = \frac{1}{{tan}^{- 1} t}$ $1/x=1/tan^(-1)t$

Explanation:

$\frac{1}{x} = \frac{1}{{tan}^{- 1} t}$ $1/x=1/tan^(-1)t$

Note that $\frac{1}{x} = x^{- 1}$ $1/x=x^(-1)$ does not mean that $\frac{1}{x} = tan t$ $1/x=tant$

as $x = {tan}^{- 1} t$ $x=tan^(-1)t$ means the angle $x$ $x$ , whose tangent is $t$ $t$ and this does not mean that tangent of angle $\frac{1}{x}$ $1/x$ is $\frac{1}{t}$ $1/t$ or $\frac{1}{tan t}$ $1/tant$ or any other thing apparently similar.

In short, ${tan}^{- 1} t$ $tan^(-1)t$ is different from $\frac{1}{tan t}$ $1/tant$ . The fact is that the latter is $cot t$ $cott$ , a ratio, while $x$ $x$ is an angle, which could be in degrees, radians or even Grads.

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If $x = {tan}^{- 1} t$ $x=tan^(-1)t$ , then is $\frac{1}{x} = \frac{1}{tan t}$ $1/x=1/tant$ ?

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Explanation:

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Explanation:

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If $x = {tan}^{- 1} t$ $x=tan^(-1)t$ , then is $\frac{1}{x} = \frac{1}{tan t}$ $1/x=1/tant$ ?