Question

Is cot^-1 x = tan ^-1 1/x ???

Answer 1

See below

Explanation:

`cot^-1 (x) = tan ^-1 (1/x)`

Thinking of it in terms of cotangent being adjacent leg over opposite leg and tangent being opposite leg over adjacent leg, and since we are using inverse trig we are solving for an angle:

`cot^-1 (x/1) = tan ^-1 (1/x)`

`cot^-1 (A/O) = tan ^-1 ("O"/A)`

We can see that the `1` is the opposite leg in both ratios, and `x` is the adjacent leg in both the ratios, so would the angle be the same?