# how can we derivate cosecx by first principle

Let f (x) = cosec x. Accordingly, from the first principle,

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os x/sin x = 1/(sin x/cos x) = 1/tan x = cot x and 1/sin x = csc x
$lim_{hto0},frac{frac{1}{sin(x,+,h)},-,frac{1}{sin{x}}}{h},=,lim_{hto0},frac{frac{1}{sin{x}cos{h},+,cos{x}sin{h}},-,frac{1}{sin{x}}}{h}
=,lim_{hto0},frac{sin{x},-,sin{x}cos{h},-,cos{x}sin{h}}{hsin{x}(sin{x}cos{h},+,cos{x}sin{h})},=,lim_{hto0},frac{sin{x},-,sin{x}cos{h}}{hsin{x}(sin{x}cos{h},+,cos{x}sin{h})},-,lim_{hto0},frac{cos{x}sin{h}}{hsin{x}(sin{x}cos{h},+,cos{x}sin{h})}
Rightarrow,lim_{hto0},frac{sin{x},-,sin{x}cos{h}}{hsin{x}(sin{x}cos{h},+,cos{x}sin{h})},=,lim_{hto0},frac{1,-,cos{h}}{h(sin{x}cos{h},+,cos{x}sin{h})},=,lim_{hto0},frac{1,-,cos{h}}{h},cdot,lim_{hto0},frac{1}{sin{x}cos{h},+,cos{x}sin{h}},=,0
Rightarrow,lim_{hto0},frac{cos{x}sin{h}}{hsin{x}(sin{x}cos{h},+,cos{x}sin{h})},=,lim_{hto0},frac{cos{x}}{sin{x}(sin{x}cos{h},+,cos{x}sin{h})},=,frac{cos{x}}{sin^2{x}},Rightarrow,therefore,0,-,frac{cos{x}}{sin^2{x}},=,-frac{cos{x}}{sin^2{x}},=,-csc{x}cot{x}$

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