How do I prove tan⁡1<π2? Prove that the equation sin⁡xsin⁡(sin⁡x)=π2cos⁡(sin⁡x) Let t=sin⁡x,−1≤t≤1. Then the expression...

How do I prove ${\displaystyle \tan 1<\frac{\pi }{2}}$?
Prove that the equation
${\displaystyle \sin x\sin \left(\sin x\right)=\frac{\pi }{2}\cos \left(\sin x\right)}$
Let ${\displaystyle t=\sin x,-1\le t\le 1}$. Then the expression above is equvalent to ${\displaystyle t\sin t=\frac{\pi }{2}\cos t}$. As the function ${\displaystyle f\left(t\right)=t\sin t-\frac{\pi }{2}\cos t}$ is even, and t=0 is not a solution, I have to prove that f(t) has no positive roots (t>0). So, for the left side ${\displaystyle 0<t\le 1}$ and ${\displaystyle 0<\sin t\le \sin 1}$, then ${\displaystyle t\sin t\le \sin 1}$. For the right side ${\displaystyle \cos t\ge \cos 1}$, so ${\displaystyle \frac{\pi }{2}\cos t\ge \frac{\pi }{2}\cos 1}$. The objective is to prove that ${\displaystyle \sin 1<\frac{\pi }{2}\cos 1}$, or, equivalently, ${\displaystyle \tan 1<\frac{\pi }{2}}$