What is the derivative of $y = \frac{1}{sec x - tan x}$ $y = 1/(secx- tanx)$ ?

Question

1705 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-01T16:22:24

$y' = 1/(1 - sinx)$

Try to write in sine and cosine. Use the identities $secx = 1/cosx$ and $tanx = sinx/cosx$ .

$y = 1/(1/cosx - sinx/cosx)$

$y = 1/((1 - sinx)/cosx)$

$y = cosx/(1 - sinx)$

Differentiate this using the quotient rule.

$y' = (-sinx(1 - sinx) - cosx(-cosx))/(1- sinx)^2$

$y' = (-sinx + sin^2x + cos^2x)/(1 - sinx)^2$

Use $sin^2x + cos^2x = 1$ :

$y' = (1 - sinx)/((1 -sinx)(1 - sinx))$

$y' = 1/(1 - sinx)$

Hopefully this helps!

What is the derivative of y = 1/(secx- tanx)y=1secx−tanx y = 1/(secx- tanx)?