If $x\in \left(0,\frac{\pi}{4}\right)$ then $\frac{\cos x}{(\sin^2 x)(\cos x-\sin x)}>8$ 
If $\displaystyle x\in \left(0,\frac{\pi}{4}\right)\;,$ Then prove that $\displaystyle \frac{\cos x}{\sin^2 x(\cos x-\sin x)}>8$

$\bf{My\; Try::}$ Let $$f(x) = \frac{\cos x}{\sin^2 x(\cos x-\sin x)}=\frac{\sec^2 x}{\tan^2 x(1-\tan x)} = \frac{1+\tan^2 x}{\tan^2(1-\tan x)}$$
Now Put $\tan x= t \in (0,1)\;,$ Then $$h(t) = \frac{1+t^2}{t^2(1-t)}\;\;, 0<t<1 $$  where $h(t)=f(x)$.
Now How can i solve it after that, Help Required ,Thanks

Your original derivation was correct. Now, you have to take the derivative of $\frac{1+t^2}{t^2(1-t)}$ and find when it is equal to zero. Then, take the second derivative at that point; if the second derivative is negative, then you have a minimum. Find the value of the function at that point, and if it's larger than 8, then you've proven it.
