((sin^3 x−sin x+cos x)/(sin x))=cot x −cos^2 x prove this


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Question Number 182848 by malithxd last updated on 15/Dec/22

$\frac{{s i n}^{3} x - s i n x + c o s x}{s i n x} = c o t x - {c o s}^{2} x$ $p r o v e t h i s$
	Answered by Rasheed.Sindhi last updated on 15/Dec/22

	$\frac{{s i n}^{3} x - s i n x + c o s x}{s i n x} = c o t x - {c o s}^{2} x$ $L H S : \sin^{2} x - 1 + \frac{\cos x}{\sin x}$ $= - (1 - \sin^{2} x) + \frac{\cos x}{\sin x}$ $- \cos^{2} x + \cot x = R H S$
	Answered by manxsol last updated on 15/Dec/22

	$/ s i n x$ ${s i n}^{2} - 1 + c t x$ $- {c o s}^{2} + c o t x = c o t x - {c o s}^{2} x$
	Answered by cortano1 last updated on 15/Dec/22

	$\frac{\sin^{3} x - \sin x + \cos x}{\sin x}$ $= \frac{\sin x (\sin^{2} x - 1) + \cos x}{\sin x}$ $= \sin^{2} x - 1 + \cot x$ $= - \cos^{2} x + \cot x$
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