Question #91cad

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Answer 1 · 2017-04-13T19:15:54

$2/sinx, or, 2cscx.$

In what follows, $cosx!=1 iff sinx!=0$

The Expression= $sinx/(1-cosx)+(1-cosx)/sinx$

$={sinx*sinx+(1-cosx)*(1-cosx)}/{(1-cosx)sinx}$

$={sin^2x+(1-cosx)^2}/{(1-cosx)sinx}$

$={sin^2x+1-2cosx+cos^2x}/{(1-cosx)sinx}$

$=(1+1-2cosx)/{(1-cosx)sinx}$

$=(2-2cosx)/{(1-cosx)sinx}$

$={2(1-cosx)}/{(1-cosx)sinx}$

$=2/sinx, or, 2cscx.$