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) Identity
B) Not an 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) to 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 can be rewritten using the identity $\tan^2x = \frac{\sin^2x}{\cos^2x}$, so the RHS becomes $-2\frac{\sin^2x}{\cos^2x}$.Without even graphing, by simplifying and comparing the expressions, it's clear that the LHS and RHS represent different relationships between trigonometric functions. The LHS involves a ratio of sums and differences of $\sin x$ and its reciprocal, while the RHS is a scaled square of the tangent function. These expressions do not simplify to the same form for all $x$, indicating that the given equation is not an identity.

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 $\frac { \sin x + \csc x } { \sin x - \csc x } = - 2 \tan ^ { 2 } x$