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Chapter 9: Problem 11

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

Short Answer

Expert verified
Question: Prove that \(\tan \left(180^{\circ}-\theta\right) = -\tan \theta\). Solution: To prove the identity, we first converted the angle measurement from degrees to radians. This allowed us to apply the formula for the tangent of the difference of two angles. By evaluating the tangent of pi and substituting it into the expression, we were able to simplify and show that \(\tan \left(180^{\circ}-\theta\right)=-\tan \theta\).

Step by step solution

01

Convert degrees to radians

To make it easier to work with this expression, we will first convert the angle measurement from degrees to radians. Recall that \(180^{\circ}=\pi\) radians. So, we can rewrite the given expression as: $$ \tan \left( \pi - \theta \right) = -\tan \theta $$
02

Use tangent of angle difference formula

Recall the formula for the tangent of the difference of two angles: $$ \tan (A-B) = \frac{\tan A - \tan B}{1 + \tan A\tan B} $$ In our case, \(A=\pi\) and \(B=\theta\). Let's apply this formula to our expression: $$ \tan \left(\pi - \theta\right) = \frac{\tan \pi - \tan \theta}{1 + \tan \pi\tan \theta} $$
03

Evaluate tangent of pi

Recall that the tangent of an angle is the ratio of the sine to the cosine: \(\tan x = \frac{\sin x}{\cos x}\). Let's evaluate \(\tan \pi\): $$ \tan \pi =\frac{ \sin \pi}{\cos \pi} = \frac{0}{-1} = 0 $$
04

Substitute tangent of pi in the expression

Now that we have the value of \(\tan \pi\), let's substitute it into the expression from Step 2: $$ \tan \left(\pi - \theta\right) = \frac{0 - \tan \theta}{1 + 0\tan \theta} = \frac{-\tan \theta}{1} $$
05

Simplify and conclude the proof

Finally, we can simplify our expression to obtain the desired result: $$ \tan \left(\pi - \theta\right) = -\tan \theta $$ This completes our proof, and we've shown that \(\tan \left(180^{\circ}-\theta\right)=-\tan \theta\).

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