MHB What values of tan alpha and tan beta satisfy a trigonometric inequality?

[Albert1]

$0<\alpha < \dfrac {\pi}{2}$
$0<\beta < \dfrac {\pi}{2}$
prove:
$(1): \,\, \dfrac{1}{ \cos^2 \alpha}+ \dfrac {1}{ \sin^2 \alpha \, \sin^2 \beta \, \cos^2 \beta} \geq 9 $
determine the values of $ \tan \alpha$ and $ \tan \beta $ when :
$(2): \: \dfrac{1}{ \cos^2 \alpha}+ \dfrac {1}{ \sin^2 \alpha \, \sin^2 \beta \, \cos^2 \beta} =9 $

 

[Opalg]

[sp]Let $x = \sin^2\alpha$, $y = \sin^2(2\beta) = 4\sin^2\beta\cos^2\beta.$ Then $0\leqslant x\leqslant 1$ and $0\leqslant y\leqslant 1.$ The inequality $\dfrac{1}{ \cos^2 \alpha}+ \dfrac {1}{ \sin^2 \alpha \, \sin^2 \beta \, \cos^2 \beta} \geqslant 9$ becomes $\dfrac1{1-x} + \dfrac4{xy} \geqslant 9$, or $$\dfrac4y \geqslant 9x - \frac x{1-x} = \frac{8x-9x^2}{1-x} = \frac{4(1-x) - (3x-2)^2}{1-x} = 4 - \frac{(3x-2)^2}{1-x}.$$ The left side is clearly $\geqslant4$ and the right side is clearly $\leqslant 4.$ So the inequality is satisfied, with equality only if $x = 2/3$ and $y=1$. That occurs when $\sin\alpha = \sqrt{2/3}$ (so $\tan\alpha = \sqrt2$) and $\sin(2\beta)=1$ (so $\tan\beta = 1$).[/sp]

 

[Albert1]

[sp]Let $x = \sin^2\alpha$, $y = \sin^2(2\beta) = 4\sin^2\beta\cos^2\beta.$ Then $0\leqslant x\leqslant 1$ and $0\leqslant y\leqslant 1.$ The inequality $\dfrac{1}{ \cos^2 \alpha}+ \dfrac {1}{ \sin^2 \alpha \, \sin^2 \beta \, \cos^2 \beta} \geqslant 9$ becomes $\dfrac1{1-x} + \dfrac4{xy} \geqslant 9$, or $$\dfrac4y \geqslant 9x - \frac x{1-x} = \frac{8x-9x^2}{1-x} = \frac{4(1-x) - (3x-2)^2}{1-x} = 4 - \frac{(3x-2)^2}{1-x}.$$ The left side is clearly $\geqslant4$ and the right side is clearly $\leqslant 4.$ So the inequality is satisfied, with equality only if $x = 2/3$ and $y=1$. That occurs when $\sin\alpha = \sqrt{2/3}$ (so $\tan\alpha = \sqrt2$) and $\sin(2\beta)=1$ (so $\tan\beta = 1$).[/sp]

nice solution (Yes)