In the given equation below , applying the formula for the derivative of inverse trigonometric functions , what is the ′′u ′′ from the given function. y = cosec^(−1) [ sin (((1+sin x)/(cos x)))]


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Question Number 160933 by blackmamba last updated on 09/Dec/21

$I n t h e g i v e n e q u a t i o n b e l o w, a p p l y i n g$ $t h e f o r m u l a f o r t h e d e r i v a t i v e o f$ $i n v e r s e t r i g o n o m e t r i c f u n c t i o n s,$ $w h a t i s t h e^{″} u^{″} f r o m t h e g i v e n f u n c t i o n .$ $y = {cosec}^{- 1} [\sin (\frac{1 + \sin x}{\cos x})]$
	Answered by bobhans last updated on 10/Dec/21

	$y = {cosec}^{- 1} [\sin (\frac{1 + \sin x}{\cos x})]$ $cosec y = \sin (\frac{1 + \sin x}{\cos x})$ $(- cosec y \cot y) y^{'} = (\frac{1 + \sin x}{\cos^{2} x}) \cos (\frac{1 + \sin x}{\cos x})$ $y^{'} = (\frac{1 + \sin x}{\cos^{2} x}) \tan (\frac{1 + \sin x}{\cos x}) \tan ({cosec}^{- 1} [\sin (\frac{1 + \sin x}{\cos x})])$
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