Graph the expression on each side of the equals symbol to determine whether the equation might be an identity.
-$$\frac { \sin x + \csc x } { \sin x - \csc x } = - 2 \tan ^ { 2 } x$$
A) Not an Identity
B) Identity

Question

Quizplus · Accepted Answer

To determine if the given equation is an identity, we can simplify the left-hand side (LHS) and the right-hand side (RHS) separately and see if they are equivalent for all values of $x$. The LHS is $\frac{\sin x + \csc x}{\sin x - \csc x}$. Since $\csc x = \frac{1}{\sin x}$, we can rewrite the LHS as $\frac{\sin x + \frac{1}{\sin x}}{\sin x - \frac{1}{\sin x}}$.The RHS is $-2\tan^2x$, which simplifies to $-2\left(\frac{\sin x}{\cos x}\right)^2$.Without even graphing, by simplifying both sides, it's clear that the LHS involves a ratio of sine functions and their reciprocals, while the RHS involves a squared tangent function multiplied by -2. These expressions do not simplify to the same form for all values of $x$, indicating that the equation is not an identity. Graphing both sides would further confirm that they do not produce the same graph for all values of $x$.

Graph the Expression on Each Side of the Equals Symbol sin⁡x+csc⁡xsin⁡x−csc⁡x=−2tan⁡2x\frac { \sin x + \csc x } { \sin x - \csc x } = - 2 \tan ^ { 2 } xsinx−cscxsinx+cscx​=−2tan2x

Graph the Expression on Each Side of the Equals Symbol $\frac { \sin x + \csc x } { \sin x - \csc x } = - 2 \tan ^ { 2 } x$