Question

Show that \(\arctan (\sinh x)=\arcsin (\tanh x)\).

Short answer

The given expression \(\arctan (\sinh x)=\arcsin (\tanh x)\) can be proved by converting the functions to their exponential equivalents and applying proper identities. After performing the necessary steps and comparing both sides, we can assure that the expression is true.

Step-by-step solution

Convert the Left Side

Begin by converting \(\sinh x\) into exponential form. This is given by \(\sinh x = \frac {e^x - e^{-x}}{2}\). So, \(\arctan(\sinh x) = \arctan(\frac {e^x - e^{-x}}{2})\).

Apply Arctan Identity

Now apply the identity \(\arctan (a) = \frac{\pi}{4} + \frac{1}{2}\ln(1-a^2)\) to rewrite \(\arctan(\frac {e^x - e^{-x}}{2}) = \frac{\pi}{4} + \frac{1}{2}\ln(1-(\frac {e^x - e^{-x}}{2})^2)\).

Convert the Right Side

Similar to Step 1, convert \(\tanh x\) into exponential form, i.e., \(\tanh x = \frac {e^x - e^{-x}}{e^x + e^{-x}}\). So, \(\arcsin(\tanh x) = \arcsin (\frac {e^x - e^{-x}}{e^x + e^{-x}})\).

Apply Arcsin Identity

Now apply the arcsin identity, \(\arcsin(a) = \frac{\pi}{2} - \arccos(a)\), to rewrite \(\arcsin (\frac {e^x - e^{-x}}{e^x + e^{-x}}) = \frac{\pi}{2} - \arccos(\frac {e^x - e^{-x}}{e^x + e^{-x}})\).

Compare Both Sides and Simplify

After performing the substitutions and identity applications, both sides of the equation should be compared. If the left side equals the right side, then the original expression \(\arctan (\sinh x)=\arcsin (\tanh x)\) holds true.