Question

If $f\left(x\right)=\frac{\sec x+\tan x-1}{\tan x-\sec x+1}$, then find $f^{\prime }\left(\frac{\pi }{3}\right)$

Answer 1

$secx=\frac{1+ta{n}^2\frac{x}{2}}{1-ta{n}^2\frac{x}{2}}$

$tanx=\frac{2tan\frac{x}{2}}{1-ta{n}^2\frac{x}{2}}$

$f\left(x\right)=\frac{1+ta{n}^2\frac{x}{2}+2tan\frac{x}{2}-1+ta{n}^2\frac{x}{2}}{-1-ta{n}^2\frac{x}{2}+2tan\frac{x}{2}+1-ta{n}^2\frac{x}{2}}$

$f\left(x\right)=\frac{2ta{n}^2\frac{x}{2}+2tan\frac{x}{2}}{-2ta{n}^2\frac{x}{2}+2tan\frac{x}{2}}$

$f\left(x\right)=\frac{2tan\frac{x}{2}(tan\frac{x}{2}+1)}{2tan\frac{x}{2}(-tan\frac{x}{2}+1)}$

$f\left(x\right)=\frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}$

$f^{{}^{\prime }}\left(x\right)=\frac{\frac{1}{2}se{c}^2\frac{x}{2}(1-tan\frac{x}{2})+\frac{1}{2}se{c}^2\frac{x}{2}(1+tan\frac{x}{2})}{(1-tan\frac{x}{2}{)}^2}$

$f^{{}^{\prime }}\left(x\right)=\frac{se{c}^2\frac{x}{2}}{(1-tan\frac{x}{2}{)}^2}$

$f^{{}^{\prime }}\left(\frac{\Pi }{3}\right)=\frac{\frac{4}{3}}{\frac{4-2\sqrt{3}}{3}}$