# Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 14

Show that $\displaystyle \tan h (x + y) = \frac{\tan h x + \tan hy}{1 + \tan h x \tan hy}$

Solving for the left-hand side of the equation

Using Hyperbolic Function

\begin{aligned} \tan hx =& \frac{\sin hx}{\cos hx} = \frac{e^x - e^{-x}}{e^x + e^{-x}} \\ \\ \tan h (x + y) =& \frac{e^{(x + y)} - e^{-(x + y)} }{e^{(x + y)} + e^{-(x + y)}} \\ \\ \tan h (x + y) =& \frac{e^x e^y - e^{-x} e^{-y}}{e^x e^y + e^{-x} e^{-y}} \end{aligned}

Using Hyperbolic Identities

\begin{aligned} & \cos hx + \sin hx = e^x \text{ and } \cos hx - \sin hx = e^{-x} \\ \\ & \tan h(x + y) = \frac{(\cos hx + \sin hx)(\cos hy + \sin hy) - (\cos hx - \sin hx)(\cos hy - \sin hy)}{(\cos hx + \sin hx)(\cos hy + \sin hy) + (\cos hx - \sin hx)(\cos hy - \sin hy)} \\ \\ & \tan h(x + y) = \frac{\cos hx \cos hy + \cos hx \sin hy + \sin hx \cos hy + \sin hx \sin hy - (\cos hx \cos hy - \cos hx \sin hy)(- \sin hx \cos hy + \sin hx \sin hy)}{\cos hx \cos hy + \cancel{\cos hx \sin hy} + \cancel{\sin hx \cos hy} + \sin hx + \sin hy + \cos hx \cos hy - \cancel{\cos hx \sin hy} - \cancel{\sin hx \cos hy} + \sin hx \sin hy} \\ \\ & \tan h(x + y) = \frac{\cancel{\cos hx \cos hy} + \cos hx \sin hy + \sin hx \cos hy + \cancel{\sin hx \sin hy} - \cancel{\cos hx \cos hy} + \cos hx \sin hy + \sin hx \cos hy - \cancel{\sin hx \sin hy}}{2 \cos hx \cos hy + 2 \sin hx \sin hy} \\ \\ & \tan h(x + y) = \frac{2 \cos hx \sin hy + 2 \sin hx \cos hy}{2 \cos hx \cos hy + 2 \sin hx \sin hy} \\ \\ & \tan h(x + y) = \frac{\cancel{2} (\cos hx \sin hy + \sin hx \cos hy)}{\cancel{2} (\cos hx \cos hy + \sin hx \sin hy)} \\ \\ & \tan h(x + y) = \frac{\cos hx \sin hy + \sin hx \cos hy}{\cos hx \cos hy + \sin hx \sin hy} \cdot \frac{\displaystyle \frac{1}{\cos hx \cos hy}}{\displaystyle \frac{1}{\cos hx \cos hy}} \\ \\ & \tan h(x + y) = \frac{\displaystyle \frac{\cancel{\cos hx} \sin hy}{\cancel{\cos hx} \cos hy} + \frac{\sin hx \cancel{\cos hy}}{\cos hx \cancel{\cos hy}}}{\displaystyle \frac{\cancel{\cos hx \cos hy}}{\cancel{\cos hx \cos hy}} + \frac{\sin hx \sin hy}{\cos hx \cos hy}} \\ \\ & \tan h(x + y) = \frac{\displaystyle \frac{\sin hy}{\cos hy} + \frac{\sin hx}{\cos hx}}{\displaystyle 1 + \frac{\sin hx}{\cos hx} \cdot \frac{\sin hy}{\cos hy}} \\ \\ & \tan h(x + y) = \frac{\tan hy + \tan hx}{1 + \tan hx \tan hy} \\ \\ & \text{or} \\ \\ & \tan h(x + y) = \frac{\tan hx + \tan hy}{1 + \tan hx \tan hy} \end{aligned}

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