Determine if the following is an identity.
-$$\frac { \cot ^ { 2 } x } { \csc x - 1 } = \frac { 1 + \sin x } { \sin x }$$
A) Identity
B) Not an identity

Question

Quizplus · Accepted Answer

The given expression is an identity. This can be shown by simplifying the left side of the equation using trigonometric identities. Recall that $\cot^2x = \csc^2x - 1$ and $\csc x = \frac{1}{\sin x}$. Substituting these into the left side, we get $\frac{\csc^2x - 1}{\csc x - 1}$. Factoring the numerator as a difference of squares, we get $\frac{(\csc x + 1)(\csc x - 1)}{\csc x - 1}$, which simplifies to $\csc x + 1 = \frac{1}{\sin x} + 1 = \frac{1 + \sin x}{\sin x}$, matching the right side of the equation.

Determine If the Following Is an Identity cot⁡2xcsc⁡x−1=1+sin⁡xsin⁡x\frac { \cot ^ { 2 } x } { \csc x - 1 } = \frac { 1 + \sin x } { \sin x }cscx−1cot2x​=sinx1+sinx​

Determine If the Following Is an Identity $\frac { \cot ^ { 2 } x } { \csc x - 1 } = \frac { 1 + \sin x } { \sin x }$