How do you verify the identity $[\frac{cos x cot x}{1} - sin x] - 1 = csc x$ $[(cosx cotx)/1-sin x]-1=cscx$ ?

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Answer 1 · 2015-06-12T17:09:19

Is there a typo here? These are not equal.

With $cot x = \frac{cos x}{sin x}$ $cotx = cosx/sinx$ and $csc x = \frac{1}{sin x}$ $cscx = 1/sinx$ :

$cos x cot x - sin x - 1 = csc x$ $cosxcotx - sinx - 1 = cscx$

$\frac{{cos}^{2} x}{sin x} - sin x - 1 = csc x$ $cos^2x/sinx - sinx - 1 = cscx$

$\frac{{cos}^{2} x}{sin x} - sin x - 1 = \frac{1}{sin x}$ $cos^2x/sinx - sinx - 1 = 1/sinx$

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

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

${cos}^{2} x \neq [sin x + 2 {sin}^{2} x] + {cos}^{2} x$ $cos^2x ne [sinx + 2sin^2x] + cos^2x$

How do you verify the identity [cosxcotx1−sinx]−1=cscx[(cosx cotx)/1-sin x]-1=cscx?