Using $\displaystyle |\sin x|=\left\{\begin{matrix}
\displaystyle \sin x \;,& \displaystyle 0 \leq x\leq \frac{\pi}{2} \\\\
\displaystyle \sin x \;,& \displaystyle \frac{\pi}{2} \leq x\leq \pi \\\\
\displaystyle -\sin x\;, & \displaystyle\pi \leq x\leq \frac{3\pi}{2} \\\\
\displaystyle -\sin x \;,& \displaystyle\frac{3\pi}{2} \leq x\leq 2\pi
\end{matrix}\right.$ and $\displaystyle |\cos x|=\left\{\begin{matrix}
\displaystyle \cos x \;,& \displaystyle 0 \leq x\leq \frac{\pi}{2} \\\\
\displaystyle -\cos x \;,& \displaystyle \frac{\pi}{2} \leq x\leq \pi \\\\
\displaystyle -\cos x \;,& \displaystyle\pi \leq x\leq \frac{3\pi}{2} \\\\
\displaystyle \cos x \;,& \displaystyle\frac{3\pi}{2} \leq x\leq 2\pi
\end{matrix}\right.$
So $$\displaystyle f(x)=\frac{\sin x}{\sqrt{1+\tan^2x}}-\frac{\cos x}{\sqrt{1+\cot^2x}} = \sin x\cdot \left|\cos x\right|-\cos x\cdot \left|\sin x\right|$$
Here Function $f(x)$ is Periodic With Time Period $= 2\pi.$
So we will Calculate for Only one Time Period.
In $\bullet \displaystyle \; 0 \leq x\leq \frac{\pi}{2}\;,$ We get $f(x) = \sin x\cdot \cos x-\cos x\cdot \sin x = 0$
In $\bullet \displaystyle \; \frac{\pi}{2} \leq x\leq \pi\;,$ We get $f(x)=-\sin 2x .$ So $0 \leq f(x)\leq 1$
In $\bullet \displaystyle \; \pi \leq x\leq \frac{3\pi}{2}\;,$ We get $f(x)=0 .$
In $\bullet \displaystyle \; \frac{3\pi}{2} \leq x\leq 2\pi\;,$ We get $f(x)=\sin 2x .$ So $-1 \leq f(x)\leq 0$
So Here We get $\displaystyle -1 \leq f(x)\leq 1$
Best Answer
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 $\tan y?$