Question

c.) tan(-420°)=_____
-420 + 360 = -60
-60 + 360 = 300°q4
θ = 360 - 300° = 60°

Explanation:

Step1: Use the property of tangent function

\(\tan(-\alpha)=-\tan\alpha\), so \(\tan(-420^{\circ}) = -\tan420^{\circ}\)

Step2: Use the periodicity of tangent function

Since the period of \(y = \tan x\) is \(180^{\circ}\), and \(420^{\circ}=360^{\circ}+60^{\circ}\), then \(\tan420^{\circ}=\tan(360^{\circ} + 60^{\circ})\)
According to \(\tan(x + k\cdot180^{\circ})=\tan x,k\in Z\), when \(k = 2,x = 60^{\circ}\), we have \(\tan(360^{\circ}+60^{\circ})=\tan60^{\circ}\)

Step3: Calculate the value

We know that \(\tan60^{\circ}=\sqrt{3}\), so \(-\tan420^{\circ}=-\sqrt{3}\)

Answer:

\(-\sqrt{3}\)