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

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Answer 1 · 2018-06-18T12:58:04

$\frac{d}{d x} (\frac{1}{tan x} - \frac{1}{cot x}) = - {csc}^{2} (2 x)$ $d/dx( 1/tanx -1/cotx ) = -csc^2 (2x)$

Note that:

$\frac{1}{tan x} - \frac{1}{cot x} = cot x - tan x$ $1/tanx -1/cotx = cotx - tanx$

$\frac{1}{tan x} - \frac{1}{cot x} = \frac{cos x}{sin x} - \frac{sin x}{cos x}$ $1/tanx -1/cotx = cosx/sinx - sinx/cosx$

$\frac{1}{tan x} - \frac{1}{cot x} = \frac{{cos}^{2} x - {sin}^{2} x}{sin x cos x}$ $1/tanx -1/cotx = (cos^2x-sin^2x)/(sinxcosx)$

$\frac{1}{tan x} - \frac{1}{cot x} = \frac{1}{2} \frac{cos 2 x}{sin 2 x}$ $1/tanx -1/cotx = 1/2 (cos2x)/(sin2x)$

$\frac{1}{tan x} - \frac{1}{cot x} = \frac{1}{2} cot (2 x)$ $1/tanx -1/cotx = 1/2 cot(2x)$

so:

$\frac{d}{d x} (\frac{1}{tan x} - \frac{1}{cot x}) = - {csc}^{2} (2 x)$ $d/dx( 1/tanx -1/cotx ) = -csc^2 (2x)$