Explain this equality: − 90 ∘ + tan − 1 ⁡ X C R = − tan − 1 ⁡ R X C

Let arctan ⁡ 1 x = θ 1 x = tan ⁡ θ x = cot ⁡ θ = tan ⁡ ( π 2 − θ ) arctan ⁡ x = π 2 − θ arctan ⁡ x = π 2 − θSo, arctan ⁡ x = π 2 − arctan ⁡ 1 x

Find out the answer to "Explain this equality: − 90 ∘ + tan..."

veneciasp

2022-07-06

Explain this equality:

- 90^{\circ} + \tan^{- 1} \frac{X_{C}}{R} = - \tan^{- 1} \frac{R}{X_{C}}

Franco Cohen

Beginner2022-07-07Added 8 answers

Let \arctan \frac{1}{x} = θ

\frac{1}{x} = \tan θ

x = \cot θ = \tan (\frac{π}{2} - θ)

\arctan x = \frac{π}{2} - θ

\arctan x = \frac{π}{2} - θ

So,

\arctan x = \frac{π}{2} - \arctan \frac{1}{x}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?