Inequality.

[man111]

If $0<x<\frac{\pi}{4}$.then prove that $\frac{cosx}{sin^2x.(cosx-sinx)}>8$

[paha]

Hint.

$$ \sin x=\frac{2t}{1+t^2},\quad \cos x=\frac{1-t^2}{1+t^2},\quad t\in [0;\tan\frac{\pi}{8}] $$

[ewert]

paha в сообщении #406353 писал(а):

Hint.

$$ \sin x=\frac{2t}{1+t^2},\quad \cos x=\frac{1-t^2}{1+t^2},\quad t\in [0;\tan\frac{\pi}{8}] $$

Жестоко. На самом деле $\dfrac{t^2+1}{t^2(1-t)}>8$ при условии, что $t\in(0;1)$, где $t=\tg x$.

[man111]

I have got it.
Thanks to all for nice solution