Use the sum or difference formula for tangent to determine the exact value of tan

Use the sum or difference formula for tangent to determine the exact value of tan(195deg ). Make sure to simplify your answer. Use the sum or difference formula for tangent to determine the exact value of tan(195) Make sure to simplify your answer.

The Correct Answer and Explanation is:

To determine the exact value of tan⁡(195∘)\tan(195^\circ)tan(195∘), we will use the difference formula for the tangent of angles, which is:tan⁡(A−B)=tan⁡(A)−tan⁡(B)1+tan⁡(A)⋅tan⁡(B)\tan(A – B) = \frac{\tan(A) – \tan(B)}{1 + \tan(A) \cdot \tan(B)}tan(A−B)=1+tan(A)⋅tan(B)tan(A)−tan(B)​

We can express 195∘195^\circ195∘ as 195∘=180∘+15∘195^\circ = 180^\circ + 15^\circ195∘=180∘+15∘, so we will apply the difference formula with A=180∘A = 180^\circA=180∘ and B=15∘B = 15^\circB=15∘.

Step 1: Apply the Difference Formula

tan⁡(195∘)=tan⁡(180∘+15∘)\tan(195^\circ) = \tan(180^\circ + 15^\circ)tan(195∘)=tan(180∘+15∘)

Using the difference formula for tangent:tan⁡(195∘)=tan⁡(180∘)−tan⁡(15∘)1+tan⁡(180∘)⋅tan⁡(15∘)\tan(195^\circ) = \frac{\tan(180^\circ) – \tan(15^\circ)}{1 + \tan(180^\circ) \cdot \tan(15^\circ)}tan(195∘)=1+tan(180∘)⋅tan(15∘)tan(180∘)−tan(15∘)​

Step 2: Simplify

From trigonometric identities, we know:tan⁡(180∘)=0\tan(180^\circ) = 0tan(180∘)=0

Therefore, the equation simplifies to:tan⁡(195∘)=0−tan⁡(15∘)1+0⋅tan⁡(15∘)\tan(195^\circ) = \frac{0 – \tan(15^\circ)}{1 + 0 \cdot \tan(15^\circ)}tan(195∘)=1+0⋅tan(15∘)0−tan(15∘)​

This simplifies further to:tan⁡(195∘)=−tan⁡(15∘)\tan(195^\circ) = -\tan(15^\circ)tan(195∘)=−tan(15∘)

Step 3: Find the Exact Value of tan⁡(15∘)\tan(15^\circ)tan(15∘)

We can use the tangent subtraction formula for 15∘=45∘−30∘15^\circ = 45^\circ – 30^\circ15∘=45∘−30∘. The formula for the tangent of a difference is:tan⁡(A−B)=tan⁡(A)−tan⁡(B)1+tan⁡(A)⋅tan⁡(B)\tan(A – B) = \frac{\tan(A) – \tan(B)}{1 + \tan(A) \cdot \tan(B)}tan(A−B)=1+tan(A)⋅tan(B)tan(A)−tan(B)​

Substitute A=45∘A = 45^\circA=45∘ and B=30∘B = 30^\circB=30∘:tan⁡(15∘)=tan⁡(45∘)−tan⁡(30∘)1+tan⁡(45∘)⋅tan⁡(30∘)\tan(15^\circ) = \frac{\tan(45^\circ) – \tan(30^\circ)}{1 + \tan(45^\circ) \cdot \tan(30^\circ)}tan(15∘)=1+tan(45∘)⋅tan(30∘)tan(45∘)−tan(30∘)​

Using known values for the tangents:tan⁡(45∘)=1,tan⁡(30∘)=13\tan(45^\circ) = 1, \quad \tan(30^\circ) = \frac{1}{\sqrt{3}}tan(45∘)=1,tan(30∘)=3​1​

Substituting these values into the formula:tan⁡(15∘)=1−131+1⋅13\tan(15^\circ) = \frac{1 – \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}}tan(15∘)=1+1⋅3​1​1−3​1​​

Simplify the numerator and denominator:tan⁡(15∘)=3−133+13\tan(15^\circ) = \frac{\frac{\sqrt{3} – 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}}tan(15∘)=3​3​+1​3​3​−1​​

Canceling out 13\frac{1}{\sqrt{3}}3​1​:tan⁡(15∘)=3−13+1\tan(15^\circ) = \frac{\sqrt{3} – 1}{\sqrt{3} + 1}tan(15∘)=3​+13​−1​

Step 4: Conclusion

Since tan⁡(195∘)=−tan⁡(15∘)\tan(195^\circ) = -\tan(15^\circ)tan(195∘)=−tan(15∘), we get:tan⁡(195∘)=−3−13+1\tan(195^\circ) = -\frac{\sqrt{3} – 1}{\sqrt{3} + 1}tan(195∘)=−3​+13​−1​

This is the exact value of tan⁡(195∘)\tan(195^\circ)tan(195∘).