Question Number 162326 by stelor last updated on 28/Dec/21
![Hello please show it... a ∈ [0 , (π/4)] a ≤tan a ≤ 2a](Q162326.png)
$${Hello}\:{please}\:{show}\:{it}... \\ $$$$\:{a}\:\in\:\left[\mathrm{0}\:,\:\frac{\pi}{\mathrm{4}}\right]\:\:\:\:\:\:\:\:{a}\:\leqslant{tan}\:{a}\:\leqslant\:\mathrm{2}{a} \\ $$
Answered by mindispower last updated on 28/Dec/21
![1≤1+tg^2 (a)≤2,∀a∈[0,(π/4)] ∫_0 ^x 1da≤∫_0 ^x (1+tg^2 (a))da≤2∫_0 ^x da ⇔x≤tg(x)≤2x](Q162339.png)
$$\mathrm{1}\leqslant\mathrm{1}+{tg}^{\mathrm{2}} \left({a}\right)\leqslant\mathrm{2},\forall{a}\in\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\right] \\ $$$$\int_{\mathrm{0}} ^{{x}} \mathrm{1}{da}\leqslant\int_{\mathrm{0}} ^{{x}} \left(\mathrm{1}+{tg}^{\mathrm{2}} \left({a}\right)\right){da}\leqslant\mathrm{2}\int_{\mathrm{0}} ^{{x}} {da} \\ $$$$\Leftrightarrow{x}\leqslant{tg}\left({x}\right)\leqslant\mathrm{2}{x} \\ $$
Commented by stelor last updated on 28/Dec/21
$${thanks}.. \\ $$
Commented by mindispower last updated on 28/Dec/21
$${pleasur}\:{sir} \\ $$