Question

Given the function y = 4 - 2 sec x + tan x Find the equation of the tangent line to the curve at the point P(0,?).

Answer 1

The given curve is $y=4-2\sec x+\tan x$ .

When $x=0$ , $y(0)=4-2\sec 0+\tan 0=2$ .

So, we have to find the equation of tangent line at .

The slope of the tangent is

$\frac{\mathrm{d} y}{\mathrm{d} x}=-2\sec x \tan x+\sec^2x\\ \frac{\mathrm{d} y}{\mathrm{d} x}|_{x=0}=-2\sec 0 \tan 0+\sec^2 0=1$

The equation of the tangent is

$\frac{y-2}{x-0}=1\\ {\color{Blue} y=x+2}$