Show that $\sin\theta \tan\theta <2(1-\cos 3\theta )$ 
Show that $$\sin\theta \tan\theta <2(1-\cos 3\theta)$$ for $0<\theta<\pi/2$

MY ATTEMPT :
Let $$ t = \cos \theta $$
$$t \times E= (1-t^2) -2t(1-4t^3+3t) $$
$$t \times E= 8t^4-7t^2-2t+1 $$
After this, I'm confused about how to proceed !
Any help ? thank you
 A: As peter's comment explains, your inequality is false because $\sin(\theta)\tan(\theta)$ grows arbitrarily large as $\theta$ approaches $\frac{\pi}{2}$ from the left, while the RHS of the inequality does not.
You can show this by recalling that $\cos(x)$ is bounded between $-1$ and $1$ for all $x \in \mathbb{R}$, in particular, this means you get
\begin{align}
&1 \ge \cos(3\theta) \ge -1 \ \implies \ -1 \le -\cos(3\theta) \le 1 \ \\
\implies \ &0 \le 1-\cos(3\theta) \le 2  \ \implies \  0 \le 2\left(1-\cos(3\theta)\right) \le 4
\end{align}
so the RHS of your inequality is always bounded between $0$ and $4$.
As for why the LHS blows up, peter again explains that the $\cos(\theta)$ in the denominator is the culprit. You can explicitly see this by taking the limit as $\theta \to \frac{\pi}{2}$ from the left. Doing this you get
$$
\lim_{\theta \to \frac{\pi}{2}^-}\sin(\theta)\tan(\theta) = \lim_{\theta \to \frac{\pi}{2}^-}\sin(\theta)\frac{\sin(\theta)}{\cos(\theta)} =\left[\lim_{\theta \to \frac{\pi}{2}^-} \sin^2(\theta) \right]\left[\lim_{\theta \to \frac{\pi}{2}^-} \frac{1}{\cos(\theta)} \right]  = (1)\left[\lim_{\theta \to \frac{\pi}{2}^-} \frac{1}{\cos(\theta)} \right]
$$
since $\sin^2\left(\frac{\pi}{2}\right) = 1$. For the latter limit, recall that since $0 < \theta < \frac{\pi}{2}$, the function $\cos(\theta)$ is positive on this interval, and since $\cos\left(\frac{\pi}{2}\right) =0$, in the latter limit your dividing $1$ by a really small positive number, which results in a really big positive number. Using this, you conclude that
$$
\lim_{\theta \to \frac{\pi}{2}^-}\sin(\theta)\tan(\theta)= \lim_{\theta \to \frac{\pi}{2}^-} \frac{1}{\cos(\theta)} = + \infty
$$
You can also visually see this if you graph the $2$ functions:

where it is clear that $\sin(\theta)\tan(\theta)$ overtakes $2\left(1-\cos(3\theta)\right)$ in the interval we're analyzing.
A: As , $\lim_{ \theta\to \pi/2-} \tan(\theta) $ goes to $+\infty $, and since $\sin(\theta) $ is bounded , so as limit tends to $\pi/2- $ , $\sin(\theta) \tan(\theta) $ goes to $ +\infty $. But the right hand side function $2 (1- \cos(3\theta)) $ is bounded on $ 0 < \theta < \pi/2 $.
So, your inequality $\sin(\theta) \tan(\theta) < 2 ( 1 - \cos(3\theta)) $ is not valid on $ 0 < \theta < \pi/2 $ .
