Question

Proving Identities

Verify the identity$\frac{\sin x-1}{\sin x+1}=\frac{-{\cos }^{2}x}{{(\sin x+1)}^2}$

Short answer

The expression$\frac{\sin x-1}{\sin x+1}=\frac{-{\cos }^{2}x}{{(\sin x+1)}^2}$is an identity.

Step-by-step solution

Step 1. Given information.

An expression$\frac{\sin x-1}{\sin x+1}=\frac{-{\cos }^{2}x}{{(\sin x+1)}^2}$

Step 2. Concept used.

Divide the provided phrase into two parts: the LHS and the RHS. Simplify each one separately and compare the results. If the two outcomes are the same, the supplied expression is an identity.

Step 3. Calculation.

Now, simplify LHS of $\frac{\sin x-1}{\sin x+1}=\frac{-{\cos }^{2}x}{{(\sin x+1)}^2}$:

$\begin{array}{l}\frac{\sin x-1}{\sin x+1}=\frac{\sin x-1}{\sin x+1}\times \frac{\sin x+1}{\sin x+1}\\ =\frac{{\sin }^{2}x-1}{{(\sin x+1)}^2}\\ \because {\sin }^{2}x-1=-{\cos }^{2}x\\ =\frac{-{\cos }^{2}x}{{(\sin x+1)}^2}\end{array}$

RHS of the equation is$\frac{\sin x-1}{\sin x+1}$

Both RHS and LHS are equal. Hence it is an identity