How do you verify the identity $[(cosx cotx)/1-sin x]-1=cscx$ ?

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Answer 1 · 2015-06-12T17:09:19

Is there a typo here? These are not equal.

With $cotx = cosx/sinx$ and $cscx = 1/sinx$ :

$cosxcotx - sinx - 1 = cscx$

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

$cos^2x/sinx - sinx - 1 = 1/sinx$

$cos^2x - sin^2x - sinx = 1$

$cos^2x - sin^2x - sinx = sin^2x + cos^2x$

$cos^2x ne [sinx + 2sin^2x] + cos^2x$

How do you verify the identity [(cosx cotx)/1-sin x]-1=cscx[(cosx cotx)/1-sin x]-1=cscx?