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Chapter 9: Problem 14

Show \(\tan \left(180^{\circ}+\theta\right)=\tan \theta\)

Short Answer

Expert verified
Question: Prove that \(\tan(180^\circ+\theta) = \tan(\theta)\). Answer: We used the angle addition formula for tangent, along with the property that \(\tan(180^\circ) = \tan(0^\circ) = 0\), to demonstrate that \(\tan(180^\circ+\theta) = \tan(\theta)\).

Step by step solution

01

Recall the angle addition formula for tangent

The angle addition formula for tangent states that: \(\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha) \tan(\beta)}\) In our case, the two angles being added are \(180^\circ\) and \(\theta\). So we'll plug these values into the formula and simplify.
02

Substitute the given angles into the formula

Substitute \(\alpha = 180^\circ\) and \(\beta = \theta\) into the angle addition formula: \(\tan(180^\circ + \theta) = \frac{\tan(180^\circ) + \tan(\theta)}{1 - \tan(180^\circ) \tan(\theta)}\)
03

Recall the value of \(\tan(180^\circ)\)

The tangent function has a period of \(180^\circ\), which means that \(\tan(180^\circ + x) = \tan(x)\). In our case, since we are dealing with \(180^\circ\), we have: \(\tan(180^\circ) = \tan(0^\circ) = 0\) Now, we will substitute this value back into our equation from Step 2.
04

Substitute the value of \(\tan(180^\circ)\) into the equation

Replace \(\tan(180^\circ)\) with 0 in our equation: \(\tan(180^\circ + \theta) = \frac{0 + \tan(\theta)}{1 - 0 \cdot \tan(\theta)}\)
05

Simplify the expression

Since the value of \(\tan(180^\circ)\) is 0, our expression becomes: \(\tan(180^\circ + \theta) = \frac{\tan(\theta)}{1}\) As dividing by one doesn't change the value, we can further simplify this to: \(\tan(180^\circ + \theta) = \tan(\theta)\) This completes the proof that \(\tan(180^\circ+\theta) = \tan(\theta)\).

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