Question

Show that the derivative of y=(1-cosx)/(sinx) equals the derivative of y=cscx-cotx.

Answer 1

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