Question

Find the exact value using asum or difference identity.tan 15°[?] - ✓-Hint: tan(A + B) =tan Atan B1 F tan A tan B

Answer 1

2 - √3

Step-by-step explanation

15 degrees is the difference between 45 degrees and 30 degrees. We know the exact value of the tangent of these angles,

`\begin{cases}\tan 30\degree=\frac{\sqrt[]{3}}{3} \\ \\ \tan 45\degree=1\end{cases}`

Using the identity,

`\tan (A-B)=(\tan A-\tan B)/(1+\tan A\tan B)`

If A is 45° and B is 30°,

`\tan (45-30)=\tan (15)=(\tan45-\tan30)/(1+\tan45\tan30)=\frac{1-\frac{\sqrt[]{3}}{3}}{1+1\cdot\frac{\sqrt[]{3}}{3}}=\frac{1-\frac{\sqrt[]{3}}{3}}{1+\frac{\sqrt[]{3}}{3}}`

Add the numbers in the denominator and the numerator,

`\tan (15)=\frac{\frac{3-\sqrt[]{3}}{3}}{\frac{3+\sqrt[]{3}}{3}}=\frac{3-\sqrt[]{3}}{3+\sqrt[]{3}}`

To simplify this expression, multiply and divide by (3 - √3),

`\tan (15)=\frac{(3-\sqrt[]{3})}{(3+\sqrt[]{3})}\cdot\frac{(3-\sqrt[]{3})}{(3-\sqrt[]{3})}`

The denominator is a difference of two squares, and the numerator is the square of the binomial,

`\tan (15)=\frac{(3-\sqrt[]{3})^2}{3^2-(\sqrt[]{3})^2}=\frac{3^2-6\sqrt[]{3}+3}{9-3}=\frac{12-6\sqrt[]{3}}{6}=(12)/(6)-\frac{6\sqrt[]{3}}{6}=2-\sqrt[]{3}`

Hence, the value of tan15° is 2 - √3.