Question

how solve this dy/dx=d.sinx.cosecx+secx.cotx/dx

Answer 1

d/dx (sinx cosecx + secx cotx )
= d/dx (sinx cosecx ) + d/dx (secx cotx)
= ( sinx d/dx (cosecx) + cosecx d/dx (sinx) ) + ( secx d/dx (cotx) + cotx d/dx (secx) )
= sinx (-cosecx cot x) + cosecx cosx + secx (-sinx) + cotx (secx tanx)
= cosecx (-sinx cotx + cosx) + secx (-sinx + cotx tanx)
{ putting cotx = cosx/sinx, we get (- cosx + cosx = 0) }
{ also putting cotx tanx = 1 }
= 0 + secx (-sinx + 1 )
= secx - secx sinx
{ putting secx sinx = tanx }
= secx - tanx