How do you find the exact value of ${tan}^{- 1} 0$ $tan^-1 0$ ?

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Answer 1 · 2017-09-28T02:21:59

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Remember that $tan (θ) = \frac{sin (θ)}{cos (θ)}$ $tan(theta) = sin(theta)/cos(theta)$

You're looking for some angle $θ$ $theta$ for which $\frac{sin (θ)}{cos (θ)} = 0$ $sin(theta)/cos(theta) = 0$

...this occurs whenever $sin (θ) = 0$ $sin(theta) = 0$ .

Which happens when $θ = 0, π, 2 π,$ $theta = 0, pi, 2pi,$ (etc.)

For the inverse trig functions, you use only the first angle in this list.

So therefore, you can deduce that ${tan}^{- 1} (0) = 0$ $tan^-1(0) = 0$

GOOD LUCK!

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