Question

Show that $f\left(x\right)=\frac{x}{1+x\tan x},xϵ(0,\frac{\pi }{2})$ is maximum when $x=\cos x$

Answer 1

$f\left(x\right)=\frac{x}{1+x\tan x}nϵ(0,\frac{n}{2})$

$f^{{}^{\prime }}\left(x\right)=\frac{-(\tan x-x\log \cos x)x+(1+x\tan x)}{{(1+x\tan x)}^2}$

$f^{{}^{\prime }}\left(x\right)=\frac{1+x\tan x-\tan x+x\log \left(\cos x\right)}{{(1+x\tan x)}^2}$

$f^{{}^{\prime }}\left(x\right)=0$

$1+x\tan x-\tan x+x\log \left(\cos x\right)=0$

$\frac{x\cos x}{\cos x+x\sin x}=(\cos x+x\sin x)(-x\sin x+\cos x)$

$\Rightarrow \frac{x\cos x(-\sin x+\sin x+x\cos x)}{{(\cos x+x\sin x)}^2}$

$\Rightarrow \frac{-x\sin x\cos x+{\cos }^{2}x-x^2{\sin }^{2}x+x\sin x\cos x-x^2{\cos }^{2}x}{{\cos }^{2}x+x^2{\sin }^{2}x+2x\cos x\sin x}$

$f^{{}^{\prime }}\left(x\right)=0$

$\Rightarrow {\cos }^{2}x-x^2=0$

$f\left(x\right)=\frac{\cos x}{1+\sin x}=\frac{(1-\sin x)}{\cos x}$

$x=1\cos x$

$f\left(x\right)=\frac{-\cos x}{1-\sin x}=\frac{-\cos x(1+\sin x)}{{\cos }^{2}x}\Rightarrow \frac{-(1+\sin x)}{\cos x}$