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let y1=(tanx)cotx
taking log on both sides, we get
logy1=cotx.logtanx
diff.w.r.to x , get
(1/y1)dy1/dx=cotx.(1/tanx).sec2x+logtanx.(-cosec2x)
dy1/dx=(tanx)cotx .cosec2x[1- logtanx]
let y2=(cotx)tanx
logy2=tanxlogcotx
1/y2.dy2/dx=tanx.(1/cotx)(-cosec2x)+logcotx.sec2x
= (cotx)tanx (sec2x)[-1+logcotx]
Hence dy/dx==(tanx)cotx .cosec2x[1- logtanx]+ (cotx)tanx (sec2x)[-1+logcotx]
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Rinkoo Gupta
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