Question

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

Answer 1

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

Replace $1\to \tan \frac{\pi }{4}$ we get

$f\left(x\right)=\frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}\tan x}$

$f\left(x\right)=\tan (\frac{\pi }{4}+x)$ using $\tan A+\tan B=\frac{\tan A+\tan B}{1-\tan A\tan B}$

Differentiating both sides w.r.t $x$ we get

$f^{\prime }\left(x\right)={\sec }^2(\frac{\pi }{4}+x)$

$f^{\prime }\left(\frac{\pi }{6}\right)={\sec }^2(\frac{\pi }{4}+\frac{\pi }{6})$

$={\sec }^2\left(\frac{3\pi +2\pi }{12}\right)$

$={\sec }^2\left(\frac{5\pi }{12}\right)$

$={\sec }^2{75}^{\circ }$

$={\left(\frac{1}{\cos ({45}^{\circ }+{30}^{\circ })}\right)}^2$

$={\left(\frac{1}{\cos {45}^{\circ }\cos {30}^{\circ }-\sin {45}^{\circ }\sin {30}^{\circ }}\right)}^2$

$={\left(\frac{1}{\frac{\sqrt{3}}{2\sqrt{2}}-\frac{\sqrt{1}}{2\sqrt{2}}}\right)}^2$

$={\left(\frac{2\sqrt{2}}{\sqrt{3}-1}\right)}^2$

$=\frac{4\times 2}{3+1-2\sqrt{3}}$

$=\frac{8}{4-2\sqrt{3}}$

$=\frac{4}{2-\sqrt{3}}$

$=\frac{4}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}$

$=\frac{4(2+\sqrt{3})}{4-3}$

$=8+4\sqrt{3}$