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on the following unit circle, ( \theta ) is in radians and ( \tan(\thet…

Question

on the following unit circle, ( \theta ) is in radians and ( \tan(\theta) = \frac{-0.99}{0.1} approx -9.9 ). image of unit circle with center at origin, x and y axes, point ( (0.1, -0.99) ) on the circle without a calculator, evaluate the following expressions to the nearest hundredth or as an exact ratio. expression: ( \tan(pi + \theta) ), ( \tan(2pi - \theta) )

Explanation:

Step1: Recall the tangent addition formula for \(\tan(\pi + \theta)\)

The tangent function has a period of \(\pi\), and the identity is \(\tan(\pi + \theta)=\tan(\theta)\) because \(\tan\) has a period of \(\pi\) (i.e., \(\tan(x + \pi)=\tan(x)\) for all \(x\) in the domain of \(\tan\)). We know that \(\tan(\theta)= - 9.9\), so \(\tan(\pi+\theta)=\tan(\theta)=-9.9\).

Step2: Recall the tangent addition formula for \(\tan(2\pi+\theta)\)

Similarly, the period of the tangent function is \(\pi\), and also, the period can be considered as \(2\pi\) (since \(\pi\) is the fundamental period, \(2\pi\) is also a period). The identity is \(\tan(2\pi+\theta)=\tan(\theta)\) because adding \(2\pi\) (a full rotation) to an angle does not change its tangent value. So \(\tan(2\pi + \theta)=\tan(\theta)=-9.9\).

Answer:

For \(\tan(\pi + \theta)\), the value is \(-9.9\).
For \(\tan(2\pi + \theta)\), the value is \(-9.9\).