Question

help please! i need help verifying trig identities. especially when they look like this: sinx/1-cosx-cotx=cscx

Answer 1

yup

Answer 2

yea

Answer 3

\[\frac{\sin x}{1 - \cos x} - \cot x?\]

Answer 4

yes

Answer 5

\[\frac{\sin x}{1 - \cos x} - \cot x\]\[= \left(\frac{\sin x}{1 - \cos x}\right)\left(\frac{1 + \cos x}{1 + \cos x} \right) - \frac{\cos x}{\sin x}\]\[= \frac{\left( \sin x \right)\left( 1 + \cos x \right)}{1 - \cos^2 x} - \frac{\cos x}{\sin x}\]\[= \frac{\left( \sin x \right)\left( 1 + \cos x \right)}{\sin^2 x} - \frac{\cos x}{\sin x}\]\[= \frac{1 + \cos x}{\sin x} - \frac{\cos x}{\sin x}\]\[= \frac{1}{\sin x} =\csc x\]

Answer 6

\[\frac{\sin x}{1-\cos x}-\cot x=\frac{\sin x}{1-\cos x}-\frac{\cos x}{\sin x}=\frac{\sin^2x-\cos x(1-\cos x)}{\sin x(1-\cos x)}\]
\[\frac{\sin ^2x-\cos x+\cos ^2x}{\sin x(1-\cos x)}=\frac{1-\cos x}{\sin x(1-\cos x)}=\frac{1}{\sin x}=\csc x\]