d. tan 37π/6\n37π/6 and π/6 differ by a multiple of 2π.\n37π/6 = 3(2π) + π/6\ncase 1, with a = π/6 and k =…

d. tan 37π/6\n37π/6 and π/6 differ by a multiple of 2π.\n37π/6 = 3(2π) + π/6\ncase 1, with a = π/6 and k = 3\nrewrite using a quotient identity since the symmetry identities are in terms of sine and cosine.\ntan 37π/6 = sin 37π/6 / cos 37π/6\n= sin(3(2π) + π/6) / cos(3(2π) + π/6)\n= sin π/6 / cos π/6 or tan π/6 quotient identity

d. tan 37π/6\n37π/6 and π/6 differ by a multiple of 2π.\n37π/6 = 3(2π) + π/6\ncase 1, with a = π/6 and k = 3\nrewrite using a quotient identity since the symmetry identities are in terms of sine and cosine.\ntan 37π/6 = sin 37π/6 / cos 37π/6\n= sin(3(2π) + π/6) / cos(3(2π) + π/6)\n= sin π/6 / cos π/6 or tan π/6 quotient identity

Answer

Explanation:

Step1: Simplify the angle

We know that $\frac{37\pi}{6}=3\times2\pi+\frac{\pi}{6}$. Since the trigonometric - functions are periodic with period $2\pi$, $\tan\frac{37\pi}{6}=\tan(3\times2\pi + \frac{\pi}{6})=\tan\frac{\pi}{6}$ because $\tan(x + 2k\pi)=\tan x$ for any real - number $x$ and integer $k$.

Step2: Use the tangent formula

The tangent function is defined as $\tan\theta=\frac{\sin\theta}{\cos\theta}$. For $\theta = \frac{\pi}{6}$, we know that $\sin\frac{\pi}{6}=\frac{1}{2}$ and $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$. So, $\tan\frac{\pi}{6}=\frac{\sin\frac{\pi}{6}}{\cos\frac{\pi}{6}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$.

Answer:

$\frac{\sqrt{3}}{3}$