How do you prove ${cos(x)cot(x)}/{1-sin(x)} - 1 = csc(x)$ ?

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Answer 1 · 2016-05-28T05:44:58

$LHS={cos(x)cot(x)}/{1-sin(x)} - 1$

$={cos(x)cot(x)}/{1-sin(x)} sinx/sinx- 1$

$=[(cosx*cosx)/cancelsinx*cancelsinx}/"(1-sinx)sinx"- 1$

$=cos^2x/"(1-sinx)sinx"- 1$

$=(1-sin^2x)/"(1-sinx)sinx"- 1$

$=((1+sinx)cancel((1-sinx)))/(cancel((1-sinx))*sinx)- 1$

$=((1+sinx)-sinx)/sinx$
$=(1+cancelsinx-cancelsinx)/sinx=1/sinx=cscx=RHS$
proved

How do you prove {cos(x)cot(x)}/{1-sin(x)} - 1 = csc(x){cos(x)cot(x)}/{1-sin(x)} - 1 = csc(x)?