What is (sectheta)/7 in terms of tantheta?

Dec 17, 2015

$\sec \frac{\theta}{7} = \left\{\begin{matrix}\frac{\sqrt{{\tan}^{2} \left(\theta\right) + 1}}{7} \text{ "-pi/2+2pik < theta < pi/2 + 2pik \\ -sqrt(tan^2(theta)+1)/7" } \frac{\pi}{2} + 2 \pi k < \theta < \frac{3 \pi}{2} + 2 \pi k\end{matrix}\right.$

Explanation:

Given the identity ${\sec}^{2} \left(\theta\right) = {\tan}^{2} \left(\theta\right) + 1$ we can take the square root of both sides of the equation to find $\sec \left(\theta\right)$ in terms of tangent. But we must account for the sign of the function.

Noting that the secant function has a period of $2 \pi$ we have

$\sec \left(\theta\right) > 0$ for $\theta \in \left(- \frac{\pi}{2} + 2 \pi k , \frac{\pi}{2} + 2 \pi k\right) , k \in \mathbb{Z}$

$\sec \left(\theta\right) < 0$ for $\theta \in \left(\frac{\pi}{2} + 2 \pi k , \frac{3 \pi}{2} + 2 \pi k\right) , k \in \mathbb{Z}$

(If we look from the standpoint of the unit circle, then the above means that the secant function is positive in quadrants $I$ and $I V$ and negative in quadrants $I I$ and $I I I$)

So, after taking the square root, we have

$\sec \left(\theta\right) = \left\{\begin{matrix}\sqrt{{\tan}^{2} \left(\theta\right) + 1} \text{ "-pi/2+2pik < theta < pi/2 + 2pik \\ -sqrt(tan^2(theta)+1)" } \frac{\pi}{2} + 2 \pi k < \theta < \frac{3 \pi}{2} + 2 \pi k\end{matrix}\right.$

Thus

$\sec \frac{\theta}{7} = \left\{\begin{matrix}\frac{\sqrt{{\tan}^{2} \left(\theta\right) + 1}}{7} \text{ "-pi/2+2pik < theta < pi/2 + 2pik \\ -sqrt(tan^2(theta)+1)/7" } \frac{\pi}{2} + 2 \pi k < \theta < \frac{3 \pi}{2} + 2 \pi k\end{matrix}\right.$