If $t=\tanh\frac{x}{2}$, prove that $\sinh x = \frac{2t}{1-t^2}$ and $\cosh x = \frac{1+t^2}{1-t^2}$. 
If $t=\tanh\frac{x}{2}$, prove that $\sinh x = \frac{2t}{1-t^2}$ and $\cosh x = \frac{1+t^2}{1-t^2}$. Hence solve the equation $7\sinh x + 20 \cosh x = 24$.

I have tried starting by writing out $\tanh\frac{x}{2}$ in exponential form and then squaring it but I can't make any progress from this.

Let $t$=$tanh$$\frac{x}{2}$.
Using Identity $sech^2$$\frac{x}{2}$=1-$t^2$
$cosh^2$$\frac{x}{2}$=$\frac{1}{1-t^2}$
$cosh$ $x$=2$cosh^2$$\frac{x}{2}$-1
$cosh$ $x$ =2 ($\frac{1}{1-t^2}$)-1
$cosh$ $x$=$\frac{1+t^2}{1-t^2}$ is obtained by simplifying the above.
Using another identity $cosh^2$$\frac{x}{2}$-$sinh^2$$\frac{x}{2}$=1
$sinh^2$$\frac{x}{2}$=$cosh^2$$\frac{x}{2}$-1
$sinh^2$$\frac{x}{2}$=($\frac{1}{1-t^2}$)-1
$sinh $$\frac{x}{2}$=$\frac{t}{\sqrt(1-t^2)}$
$sinh$ $x$=2$cosh$$\frac{x}{2}sinh$$\frac{x}{2}$
$sinh$ $x$=$\frac{2t}{1-t^2}$ is obtained by simplifying the above.
Moving to the 2nd part after replacing with $t$ in the equation, you will get
12$t^2$+17$t$-2=0.From here I guess you should solve for $t$. You will get something in terms of $\sqrt385$. What you can do is replace it in  $cosh$ $x$ and simplify by $x$ = $\underline+$$ln$($x+\sqrt{1+x^2}$). I have not tried it but still I hope it helps.
