Question

Lim x tends to pi/4 (sec^2x - 2)/tanx-1

Answer 1

you can see your answer

Answer 2

Given,

$\lim_{x \to \frac{\pi}{4}} \ \frac{\sec^{2}x - 2}{\tan x -1} \\ \\= \lim_{x \to \frac{\pi}{4}} \frac{1+\tan^{2}x - 2}{\tan x -1} \\ \\\ = \lim_{x \to \frac{\pi}{4}} \frac{tan^{2}x - 1}{\tan x - 1 } \\ \\ = \lim_{x \to \frac{\pi}{4}} \frac{(\tan x + 1)(\tan x -1)}{\tan x - 1} \\ \\ = \lim_{x \to \frac{\pi}{4}} \ {\tan x + 1} \\ \\ = 1 + 1 \\ \\ = 2$