Question

Prove that value of $\frac{x}{1+x\tan x}$ is maximum at $x=\cos x$.

Answer 1

$f\left(x\right)=\frac{x}{1+x\tan x}$

For the value to be maximum

$\frac{df\left(x\right)}{dx}=0$

So $\frac{d\left(\frac{x}{1+x+\tan x}\right)}{dx}=0$

Applying quotient rule of difrentiation

$\frac{(1+\tan x)\frac{dx}{dx}-x\frac{d}{dx}(1+x\tan x)}{{(1+x\tan x)}^2}=0or,(1+x\tan x)-x[\frac{d1}{dx}+\frac{d\left(x\tan x\right)}{dx}]=0$

[Applying product rule of differentiation]

$or,1+x\tan x-x[\frac{xd\tan x}{dx}+\tan x\frac{dx}{dx}]=0or,1+x\tan x-x^2{\sec }^{2}x-x\tan x=0or,1=x^2{\sec }^{2}xor,\frac{1}{{\sec }^{2}x}=x^2,{\cos }^{2}x=x^2\cos x=\pm x$

Hence proved