Question

If $f(x)=\frac{1+\tan x}{1-\tan x}$ then $f^{\prime }\left(\frac{\pi }{6}\right)=$

• A. $4+\sqrt{3}$
• B. $4+2\sqrt{3}$
• C. $4(2+\sqrt{3})$
• D. $(2+\sqrt{3})$

Answer 1

Answer: (C)

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