Solve the Inequality $x+\frac{x}{\sqrt{x^2-1}} \gt \frac{35}{12}$ Solve the Inequality $$x+\frac{x}{\sqrt{x^2-1}} \gt \frac{35}{12}$$
First of all the Domain of LHS is $(-\infty \:\: -1) \cup (1  \:\: \infty)$
So i assumed $x=\sec y$ since Range of $\sec y$ is $(-\infty \:\: -1) \cup (1  \:\: \infty)$
So
$$\sec y+ |\csc y| \gt \frac{35}{12}$$
Any help here to proceed?

HINT:
Clearly, we need $x>0$  so will be $\sec y,\csc y\implies0< y<\dfrac\pi2$
Now $\sec y+\csc y>\dfrac{35}{12}$
$\iff\left(\dfrac{35}{12}\right)^2<\sec^2y+\csc^2y+2\sec y\csc y=\sec^2y\csc^2y+2\sec y\csc y$  as $\sec^2y\csc^2y=\sec^2y+\csc^2y$
Set $\sec y\csc y=u$ to find $$u^2+2u>\left(\dfrac{35}{12}\right)^2\iff(u+1)^2>\left(\dfrac{37}{12}\right)^2$$
As $u>0,$  $$u+1>\dfrac{37}{12}\iff\dfrac{25}{12}<u=\dfrac2{\sin2 y}\iff\dfrac{24}{25}>\sin2y=\dfrac{2\tan y}{1+\tan^2y}$$
Can you find the range of $\tan y?$
