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 side and compare it to the right side. The left side simplifies as follows:Given: $\frac { \sin x + \csc x } { \sin x - \csc x }$Since $\csc x = \frac{1}{\sin x}$, we can rewrite the equation as: $\frac { \sin x + \frac{1}{\sin x} } { \sin x - \frac{1}{\sin x} }$Multiplying numerator and denominator by $\sin x$ to get a common denominator, we get: $\frac { \sin^2 x + 1 } { \sin^2 x - 1 }$Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can replace $\sin^2 x + 1$ with $\cos^2 x + 1$ and $\sin^2 x - 1$ with $\cos^2 x - 1$, but this does not directly lead to the right side of the equation, which is $-2\tan^2 x$.The right side, $-2\tan^2 x$, simplifies to $-2\frac{\sin^2 x}{\cos^2 x}$ using the definition of $\tan x = \frac{\sin x}{\cos x}$.Since the simplification of the left side does not directly lead to the simplification of the right side without additional manipulations that do not hold true for all values of $x$, the equation is not an identity. The simplification process shows that the two sides of the equation do not simplify to the same expression for all values of $x$, 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 $\frac { \sin x + \csc x } { \sin x - \csc x } = - 2 \tan ^ { 2 } x$