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

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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!