Question

how to verify this identity (cosxcotx)/(1-sinx)-1=cscx?

Answer 1

Remeber cotx = cosx/sinx

1/sinx = cscx

Sin^2x+cos^2x = 1

Left hand side identity = (cosxcotx)/(1-sinx) -1

= (cosx*cosx/snx)/(1-sinx) -1

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

= [(cos^2x/sinx)-1+sinx]/(1-sinx)

= [(cos^2x-sinx+sin^2x)/sinx]/(1-sinx)

= [(1-sinx)/sinx]/(1-sinx)

= (1-sinx)/sinx(1-sinx)

= 1/sinx

= cscx

= right hand side identiy.