How do you verify the identity $cscx(cscx-sinx)+(sinx-cosx)/sinx+cotx=csc^2x$ ?

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Answer 1 · 2017-01-17T20:08:22

see below

Left Hand Side:

$cscx(cscx-sinx)+(sinx-cosx)/sinx + cotx = csc^2x-cscxsinx+(sinx-cosx)/sinx + cotx$

$=csc^2x-(1/sinx)sinx+(sinx-cosx)/sinx + cosx/sinx$

$=csc^2x-(1/cancel (sinx))cancel (sinx)+(sinx-cancelcosx+ cancelcosx)/sinx$

$=csc^2x-1+sinx/sinx$

Use identity: $1+cot^2x=csc^2x$

$=cot^2x+cancelsinx/cancelsinx$

$=cot^2x+1$

$=csc^2x$

$:.=$ Right Hand Side

How do you verify the identity cscx(cscx-sinx)+(sinx-cosx)/sinx+cotx=csc^2xcscx(cscx-sinx)+(sinx-cosx)/sinx+cotx=csc^2x?