Differential: y = 1/tan x- 1/cot x ?

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Answer 1 · 2018-06-18T12:58:04

#d/dx( 1/tanx -1/cotx ) = -csc^2 (2x)#

Note that:

#1/tanx -1/cotx = cotx - tanx #

#1/tanx -1/cotx = cosx/sinx - sinx/cosx #

#1/tanx -1/cotx = (cos^2x-sin^2x)/(sinxcosx) #

#1/tanx -1/cotx = 1/2 (cos2x)/(sin2x) #

#1/tanx -1/cotx = 1/2 cot(2x) #

so:

#d/dx( 1/tanx -1/cotx ) = -csc^2 (2x)#