Sure! Let's find the tangent of [tex]\(15^\circ\)[/tex] using a detailed, step-by-step solution based on the tangent of the difference of two angles, specifically [tex]\(45^\circ\)[/tex] and [tex]\(30^\circ\)[/tex].
First, recall the tangent subtraction formula for two angles:
[tex]\[
\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}
\][/tex]
Here, we're interested in:
[tex]\[
\tan(15^\circ) = \tan(45^\circ - 30^\circ)
\][/tex]
We know that:
[tex]\[
\tan(45^\circ) = 1
\][/tex]
[tex]\[
\tan(30^\circ) = \frac{1}{\sqrt{3}}
\][/tex]
Plugging these values into the tangent subtraction formula gives:
[tex]\[
\tan(15^\circ) = \frac{\tan(45^\circ) - \tan(30^\circ)}{1 + \tan(45^\circ) \tan(30^\circ)}
\][/tex]
[tex]\[
= \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}}
\][/tex]
Let's simplify the numerator and the denominator separately:
For the numerator:
[tex]\[
1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}}
\][/tex]
For the denominator:
[tex]\[
1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}}
\][/tex]
Therefore, we have:
[tex]\[
\tan(15^\circ) = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}
\][/tex]
Since we are given the numerical values:
[tex]\[
\tan(15^\circ) ≈ 0.26794919243112264
\][/tex]
To match the format from the question:
[tex]\[
\tan 15^\circ = \sqrt{\frac{2 - \sqrt{3}}{1 + \sqrt{3}}}
\][/tex]
Breaking it down:
- Numerator of the fraction under the square root: [tex]\(\boxed{2 - \sqrt{3}}\)[/tex]
- Denominator of the fraction under the square root: [tex]\(\boxed{1 + \sqrt{3}}\)[/tex]
So the final format is:
[tex]\[
\tan 15^\circ = \sqrt{\frac{2 - \sqrt{3}}{1 + \sqrt{3}}}
\][/tex]
