Question

Expand y=log( secx+tanx) is a maclaurin series

Answer 1

Answer:

hope it's helpful for you ❤❤

Answer 2

Answer:The Maclaurin series expansion for $log(1+x)$ given above, we can write

$y = (z - z^2/2 + z^3/3 - z^4/4 + ...) - (z^2/2 - z^4/4 + z^6/6 - z^8/8 + ...)$

Step-by-step explanation:

To expand $y=log( secx+tanx)$

as a Maclaurin series , we can use the fact that the logarithm function has a Maclaurin series expansion of

$log(1 + x) = x - x^2/2 + x^3/3 - x^4/4 + ...$

for all values of x such that

$|x| < 1.$

We can rewrite $y=log( secx+tanx)$

as $y=log( 1/cos(x) + tan(x))$.

Using the identities $sec(x) = 1/cos(x) and tan(x) = sin(x)/cos(x)$

, we can rewrite this as

$y = log( (1 + sin(x))/cos(x) )$

Letting $z = sin(x)$

we can rewrite this as

$y = log( (1 + z)/cos(x) )$

Using the Maclaurin series expansion for

$log(1+x)$

given above, we can write

$y = (z - z^2/2 + z^3/3 - z^4/4 + ...) - (z^2/2 - z^4/4 + z^6/6 - z^8/8 + ...)$

This is the Maclaurin series expansion for $y=log( secx+tanx)$.

Note that the series will be valid for all values of x such that$|sin(x)| < 1$, or equivalently, such that $|x| < 90$ degrees.

for more visit-https://brainly.com/question/24188694

#SPJ2