How do you verify the identity $cosxcotx = cscx - sinx$ ?

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Answer 1 · 2018-03-17T05:47:00

All the identities you will need:
$cotx= cosx/sinx$
$cos^2x= 1-sin^2x$
$1/sinx= cscx$

Starting:
$cosxcotx= cscx-sinx$

Apply number 1 on the list:

$cosx*cosx/sinx= cscx-sinx$

Simplify:

$cos^2x/sinx= cscx-sinx$

Apply number 2 on the list:

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

Split the numerator:

$1/sinx-sin^2x/sinx= cscx-sinx$

Apply number 3 on the list:

$cscx-sinx=cscx-sinx$

Answer 2 · 2018-03-17T06:01:28

Kindly go through a Proof in the Explanation.

We have, $cosxcotx+sinx$ ,

$=cosx*cosx/sinx+sinx$ ,

$=(cos^2x+sin^2x)/sinx$ ,

$=1/sinx$ ,

$:. cosxcotx+sinx=cscx$ .

$rArr cosxcotx=cscx-sinx$ , as desired!

How do you verify the identity cosxcotx = cscx - sinxcosxcotx = cscx - sinx?