Estimating $\tan(\cos(\sin1))$ The precise question through which I came across this particular estimation is as follows:
If $x=\alpha$ is the maximum value of $x$ for which $ \left \lfloor{\sin^{-1}(\cos^{-1}(\tan^{-1}x))} \right \rfloor = 1$, then $\alpha$ lies in the interval:
A) $\left(0,\dfrac{1}{\sqrt3} \right)$
B) $\left(\dfrac{1}{\sqrt3},1 \right)$
C) $\left(1,\sqrt3 \right)$
D) $(\sqrt3, \infty)$

Finding the value of $\alpha$ isn't hard. The part where I'm stuck in is the estimation of its interval.
$$ \begin{align} &\ \left \lfloor{\sin^{-1}(\cos^{-1}(\tan^{-1}x))} \right \rfloor = 1 \\
  \implies &\ \sin^{-1}(\cos^{-1}(\tan^{-1}x)) \in \left[1,\frac{\pi}{2}\right] \\
\implies &\ \cos^{-1}(\tan^{-1}x) \in \left[\sin1,1\right] \\
\implies &\ \tan^{-1}x \in [\cos1,\cos(\sin1)] \\
\implies &\ x \in [\tan(\cos1),\tan(\cos(\sin1))] \end{align} $$
From here, it's very clear that $\alpha =\tan(\cos(\sin1))$ but now I am supposed to predict the interval in which it lies. The options seem to represent the numerical values of $\tan0, \tan\frac{\pi}{6},\tan\frac{\pi}{4},\tan\frac{\pi}{3},\tan\frac{\pi}{2} $.
So the problem essentially comes down to predicting whether $\cos(\sin1)$ lies in $\left(0, \frac{\pi}{6}\right), \left(\frac{\pi}{6}, \frac{\pi}{4}\right),\left(\frac{\pi}{4}, \frac{\pi}{3}\right), \left(\frac{\pi}{3}, \frac{\pi}{2}\right)  $
Through rough estimation, I was able to arrive at (B), which is also the correct answer.
$$  \frac{\pi}{4} < 1 < \frac{\pi}{3}  $$
$$ \frac{1}{\sqrt2} < \sin1 < \frac{\sqrt3}{2} $$
$$ \begin{align} \cos(\sin1) &\ \in (\cos0.866, \cos0.707) \\
&\ \in (\cos49.6^{\circ}, \cos40.5^{\circ}) \\
&\ \approx \cos45^{\circ} \\
&\ = \frac{1}{\sqrt2} \in \left(\frac{\pi}{6}, \frac{\pi}{4}\right) \end{align}$$
I guess my approximations were a bit too rough. But anyway, what I am looking for is a concrete mathematical way to prove that $\cos(\sin1)$ lies in the interval $\left(\frac{\pi}{6}, \frac{\pi}{4}\right)$ without any inconclusive approximations.
 A: I think this might be what you're looking for.
$$\sin(1)\approx\sin(57^\circ)\approx\sin(60^\circ)$$
Infact, this approximation has an error of $3\%$. (Keep in mind that this approximation overestimates.)
Then, you see our final expression becomes $\cos(\sqrt{3}/2)$, i.e. $\cos(0.866)$. Notice that
$$\dfrac{\pi}{4}<0.866<\dfrac{\pi}{3}$$
implying ($E$ is our expression $\cos(\sin 1)$).
$$\dfrac{\sqrt{2}}{2}>E>\dfrac12$$
Now, you notice that $\dfrac12\approx\dfrac{\pi}{6}$ (error $4\%$). Also, $\pi/4\approx 0.78$ so you're sure that $E<\pi/4$.
This was essentially your problem. We don't know if $E$ be less than $\pi/6$, being greater than $1/2$.
Well, in such cases, either you can just look at the options and mark your answer. Or you look at your inequalities and see that $0.866$ lies somewhere in between and not close to one bound. So, your final inequality, $E$ must also be in between and so it isn't harm to replace the bound by something just greater than it.
Or you can also see that you overestimated in your first line and so in the final one, the actual quantity should be slightly closer to $\sqrt{2}/2$.
Hope this helps. Ask anything if not clear :)
