I've been trying to show that $$\ln(\sec 3t+\tan 3t)=2\tanh^{-1}(\tan(3t/2))$$

I used the identity $$\tanh^{-1}x=\frac12\ln\frac{1+x}{1-x}$$ to write: $$\begin{array}{tcl} 2\tanh^{-1}(\tan(3t/2))&=&2\cdot\frac12\ln\dfrac{1+\tan(3t/2)}{1-\tan(3t/2)}\\ 2\tanh^{-1}(\tan(3t/2))&=&\ln\dfrac{1+\tan(3t/2)}{1-\tan(3t/2)}\\ \end{array}$$ Then I used $$\tan\frac{\theta}{2}=\frac{1-\cos \theta}{\sin\theta}$$ to write: $$\begin{array}{tcl} 2\tanh^{-1}(\tan(3t/2))&=&\ln\dfrac{1+\dfrac{1-\cos3t}{\sin3t}}{1-\dfrac{1-\cos 3t}{\sin 3t}}\\ 2\tanh^{-1}(\tan(3t/2))&=&\ln\dfrac{\sin 3t+1-\cos 3t}{\sin 3t-1+\cos 3t} \end{array}$$

Now, in order to show that $$\ln(\sec 3t+\tan 3t)=\ln\dfrac{\sin 3t+1-\cos 3t}{\sin 3t-1+\cos 3t}$$ I need to show that $$\sec 3t+\tan 3t=\dfrac{\sin 3t+1-\cos 3t}{\sin 3t-1+\cos 3t}$$ Now, $$\begin{array}{tcl} \sec3t+\tan3t&=&\dfrac{1}{\cos 3t}+\dfrac{\sin 3t}{\cos 3t}\\ \sec3t+\tan3t&=&\dfrac{1+\sin 3t}{\cos 3t} \end{array}$$

Now, I am stuck. How do I show that: $$\frac{\sin 3t+1-\cos 3t}{\sin 3t-1+\cos 3t}=\frac{1+\sin 3t}{\cos 3t}$$ Here is some Matlab code that seemingly verifies the identity.

plot(t,y1,'b','LineWidth',4),hold on
hold off

And the image. enter image description here


You're almost there. To prove $$\frac{\sin 3t+1-\cos 3t}{\sin 3t-1+\cos 3t}=\frac{1+\sin 3t}{\cos 3t},$$ you need $$(\sin 3t+1-\cos 3t)\cos 3t=(\sin 3t-1+\cos 3t)(1+\sin 3t).$$ Expand it: $$\sin 3t\cos 3t+\cos 3t-\cos^2 3t=\sin 3t-1+\cos 3t+\sin^2 3t-\sin 3t+\cos 3t\sin 3t,$$ which can be simplified to $$\cos^23t+\sin^2 3t=1.$$

  • $\begingroup$ I'd like to thank your for your assistance $\endgroup$ – David Aug 28 '18 at 16:42

You want to prove the identity $$\ln(\sec 3t+\tan 3t)=2\tanh^{-1}(\tan(3t/2))$$

We differentiate both sides and show that the derivatives are the same.

Note that $$ \frac {d}{dt} \ln(\sec 3t+\tan 3t) = 3 \sec 3t$$

Also we have $$ \frac {d}{dt} 2\tanh^{-1}(\tan (3t/2))$$

$$=\frac {2\sec ^ 2 ( 3t/2)}{1-\tan ^2 (3t/2)} (3/2)=3 \sec 3t$$

Thus the differennce of the two functions is a constant.

Upon evaluating the functions at $t=0$ we obtain the identity.

  • $\begingroup$ I'd like to thank your for your assistance $\endgroup$ – David Aug 28 '18 at 16:43
  • $\begingroup$ Thank for your attention and understanding. $\endgroup$ – Mohammad Riazi-Kermani Aug 28 '18 at 18:51

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