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  4. If f(x) = (1 + tan x/1 - tan x) then f' ((π/6) ) =

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Q. If $f(x) = \frac{1 + \tan \, x}{1 - \tan \,x} $ then $f' \left(\frac{\pi}{6} \right) = $

Limits and Derivatives

Solution:

If $f(x) = \frac{1 + \tan \, x}{1 - \tan \,x} $ Then
$f' (x) = \frac{\frac{d(1 + \tan \, x)}{dx} . (1 -\tan \, x) - ( 1 + \tan \, x) . \frac{d(1 - \tan \, x)}{dx}}{( 1 -\tan \,x)^2} $
$= \frac{\left(0 +\sec^{2} x\right)\left(1 -\tan x\right) - \left(1 + \tan x \right)\left(0 -\sec^{2} x\right)}{\left( 1-\tan x\right)^{2}}$
$ =\frac{2 \sec^{2} x}{1+ \tan^{2} x -2 \tan x } = \frac{2 \sec^{2} x}{\sec^{2} x - 2 \tan x} = \frac{2}{1-\sin 2x} $
$\therefore $ Where $x = \frac{\pi}{6}, f' \left(\frac{\pi}{6}\right) = \frac{2}{1-\sin \frac{\pi}{3}} = \frac{2}{1- \frac{\sqrt{3}}{2}} $
$ = \frac{4}{2-\sqrt{3}} =4\left(2 +\sqrt{3}\right)$