Question

evaluate y= ( 1-tanx /1+tanx ) then show that dy/dx= -2 /1+sin 2x
​

Answer 1

Step-by-step explanation:

We have,

$y = \frac{1 - \tan(x) }{1 + \tan(x) }$

$y = \frac{ \tan( \frac{\pi}{4} ) - \tan(x) }{1 + \tan( \frac{\pi}{4} ) . \tan(x) }$

$\implies \: y = \tan( \frac{\pi}{4} - x )$

Differentiating both sides, we get,

$\frac{dy}{dx} = - \sec ^{2} ( \frac{\pi}{4} - x)$

$\frac{dy}{dx} = - \frac{1}{ \cos^{2} ( \frac{\pi}{4} - x ) }$

$\frac{dy}{dx} = - \frac{2}{1 + \cos( \frac{\pi}{2} - 2x) }$

$\frac{dy}{dx} = - \frac{2}{1 + \sin(2x) }$